Answer:
Solid sodium hydroxide dissolves in water to form an aqueous solution of ions.
NaOH(s) ⇌ Na+(aq) + OH–(aq) ΔH1 = ?
Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride.
NaOH(s) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH2 = ?
Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride.
Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH3 = ?
What is the rate constant of a reaction if rate = 1 x 10-2 (mol/L)/s, [A] is 2 M,
[B] is 3 M, m = 2, and n = 1?
Answer:
[tex]0.10 \text{ L$^2$mol$^{-2}$s$^{-1}$}[/tex]
Explanation:
The general formula for a rate law is
[tex]\text{rate} = k\text{[A]}^m \text{[B]}^{n}[/tex]
With your numbers, the rate law becomes
1.2 mol·L⁻¹s⁻¹ = k(2 mol·L⁻¹)²(3 mol·L⁻¹)¹ = k × 4 mol²L⁻² × 3 mol·L⁻¹
= 12k mol³L⁻³
[tex]\\ k = \dfrac{\text{1.2 mol $\cdot$ L$^{-1}$s$^{-1}$} }{12\text{ mol$^{3}$L}^{-3}} = \mathbf{0.10} \textbf{ L$\mathbf{^2}$mol$^{\mathbf{-2}}$s$^{\mathbf{-1}}$}[/tex]
At what temperature in K will 0.750 moles of oxygen gas occupy 10.0 L and exert 2.50 atm of pressure
Answer:
406 K.
Explanation:
The following data were obtained from the question:
Number of mole (n) = 0.750 mole
Volume (V) = 10.0 L
Pressure (P) = 2.50 atm
Temperature (T) =.?
Note: Gas constant (R) = 0.0821 atm.L/Kmol
The temperature, T can be obtained by using the ideal gas equation as follow:
PV = nRT
2.5 x 10 = 0.75 x 0.0821 x T
Divide both side by 0.75 x 0.0821
T = (2.5 x 10) /(0.75 x 0.0821 )
T = 406 K.
Therefore, the temperature is 406 K.
Answer: 406 K
Explanation:
We can rewrite the ideal gas law to solve for T:
PV = nRT
T=PV / nR
We are given the following from the problem:
n=0.750 mol P=2.50 atm V=10.0 L
Plugging in our values and using R=0.08206 L⋅atm / K⋅mol we get:
T=(2.50 atm)(10.0 L) / (0.750 mole)(0.08206L ⋅ atm ⋅ mole K) = 406 K
Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.
Explanation:
To calculate [H3O+] in the solution we must first find the pH from the [ OH-]
That's
pH + pOH = 14
pH = 14 - pOH
To calculate the pOH we use the formula
pOH = - log [OH-]
And [OH-] = 5.5 × 10^-5 M
So we have
pOH = - log 5.5 × 10^ - 5
pOH = 4.26
Since we've found the pOH we can now find the pH
That's
pH = 14 - 4.26
pH = 9.74
Now we can find the concentration of H3O+ in the solution using the formula
pH = - log H3O+
9.74 = - log H3O+
Find the antilog of both sides
H3O+ = 1.8 × 10^ - 10 MThe solution is basic since it's pH lies in the basic region.
Hope this helps you
What is the pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid with 15.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. K a = 1.8 ×× 10-5 for CH3CO2H.
Answer:
pH = 8.72
Explanation:
This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
As this compound acts like a base, we propose this equilibrium:
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰
Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M
So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
We can avoid the quadractic equation because Kb is so small
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH = 8.72
The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
Calculation of the pH of the solution:Since the following equation should be used.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
Now
(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
So,
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
Now
Kw = Ka. Kb
Kb = Kw/Ka
And,
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵
= 5.55×10⁻¹⁰
Now
[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
Now
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH
= 8.72
Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
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The substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an:_______
A) vicinal diol
B) geminal diol
C) acetal
D) ketal
Answer:
B) geminal diol
Explanation:
Hello,
In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.
Regards.
Determine whether the following statement about equilibrium is true or false.
(a) When a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants.
(b) When a system is at equilibrium, Keq = 1.
(c) At equilibrium, the rates of the forward reaction and the reverse reaction are equal.
(d) Adding a catalyst to a reaction system will shift the position of equilibrium to the right so there are more products at equilibrium than if there was no catalyst present.
Answer:
(a) when a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants
Determining whether the statements about equilibrium is True or False
A) The concentration of the products is equal to the concentration of the reactants at equilibrium : TRUE
B) When a system is at equilibrium, Keq = 1 : TRUE
C) The rates of the forward reaction and the reverse reaction are equal at equilibrium : TRUE
D) Adding a catalyst to a reaction system will shift the position of equilibrium to the right : FALSE
Reaction at equilibriumIn a chemical reaction at equilibrium the value of Keq will be equal to 1 because the concentration of the products is equal to the concentration of the reactants in the chemica reaction. Also at equilibrium the rate of forward reaction is same as the rate of reverse reaction.
A catalyst can only affect the rate of reaction and not the amount of product ( yield of reaction).
Hence we can conclude that the answers to your questions are as listed above.
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Of the following two gases, which would you predict to diffuse more rapidly? PLZZ HELPP PLZ PLZ PLZ
Answer:
CO2 will diffuse more rapidly.
Explanation:
From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:
Rate (R) & 1/√Density (d)
R & 1/√d
But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.
Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:
Rate (R) & 1/√Molar mass (M)
R & 1/√M
From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.
Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.
This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71 g/mol
Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol
Summary
Gas >>>>>> Molar mass
Cl2 >>>>>> 71 g/mol
CO2 >>>>> 44 g/mol
From the illustration above, we can see that CO2 is lighter than Cl2.
Therefore, CO2 will diffuse more rapidly.
Answer: CO2
Explanation:
To calculate changes in concentration for a system not at equilibrium, the first step is to determine the direction the reaction will proceed. To do so, we calculate Q and compare it to the equilibrium concentration, K. We can then determine that a reaction will shift to the right if:__________
Answer:
We can then determine that a reaction will shift to the right if Q<K
Explanation:
Comparing Q with K allows to find out the status and evolution of the system:
If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium. If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium and will evolve spontaneously, decreasing the value of Qc until it equals the equilibrium constant. In this way, the concentrations of the products will decrease and the concentrations of the reagents will increase. In other words, the reverse reaction is favored to achieve equilibrium. Then the system will evolve to the left (ie products will be consumed and more reagents will be formed).If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium and will evolve spontaneously increasing the value of Qc until it equals the equilibrium constant. This implies that the concentrations of the products will increase and those of the reagents will decrease. In other words, to achieve balance, direct reaction is favored. Then the reaction will shift to the right, that is, reagents will be consumed and more products will be formed.In this case, we can then determine that a reaction will shift to the right if Q<K
There are approximately 2 × 1022 molecules and atoms in each breath we take and the concentration of CO in the air is approximately 9 ppm. How many CO molecules are in each breath we take? solution
Answer:
1.8x10¹⁷ molecules of CO are in each breath we take
Explanation:
Parts per million, ppm, is an unit of concentration in chemistry used for very diluted solutions.
A 9ppm of X in a solution means in 1 million of molecules (1x10⁶) you have only 9 molecules of X.
In a breath we have 2x10²² molecules and 9 ppm are CO. Thus, CO molecules in each breath are:
2x10²² molecules × (9 molecules CO / 1x10⁶ molecules) =
1.8x10¹⁷ molecules of CO are in each breath we take[tex]1.8\times 10^{17}[/tex] molecules of CO are in each breath we take
The calculation is as follows:A 9ppm of X in a solution represent in 1 million of molecules[tex](1\times10^6)[/tex]you have only 9 molecules of X.
Now CO molecules in each breath is
[tex]= 2\times 10^{22}\ vmolecules \times (9\ molecules\ CO \div 1\times 10^6 molecules) \\\\= 1.8\times 10^{17}[/tex]
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What are the correct half reactions for the following reaction: Cu2+ + Mg -> Cu + Mg2+
Answer:
Cu2 + 2Mg-> 2Cu+ Mg2
Explanation:
Balance the equation and make sure both the reactant and the products are the same
Hope it will be helpful
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
What is a balanced equation?A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.
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For an ideal gas condition, what is the mass (g) of N2 if the pressure is 2.0 atm, the volume is 25 mL and the temperature is 290 Kelvin.
Answer:
THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g
Explanation:
In an ideal condition
PV = nRT or PV = MRT/ MM where:
M = mass = unknown
MM =molar mass = 28 g/mol
P = pressure = 2 atm
V = volume = 25 mL = 0.025 L
R = gas constant = 0.082 L atm/mol K
T = temperature = 290 K
n = number of moles
The gas in the question is nitrogen gas
Molar mass of nitrogen gas = 14 * 2 = 28 g/mol
Then equating the variables and solving for M, we have
M = PV MM/ RT
M = 2 * 0.025 * 28 / 0.082 * 290
M = 1.4 / 23.78
M = 0.0589 g
The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g
2. Which one is the odd one
out and why?
o Water
• Hydrogen
Chlorine
o Aluminum
Answer:
Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.
Explanation:
I hope this helps bro
1. What volume in milliliters of 0.100 M HClO₃ is required to neutralize 40.0 mL of 0.140 M KOH? 2. A 25.0 mL solution of HNO₃ is neutralized with 15.7 mL of 0.250 M Ba(OH)₂. What is the concentration of the original HNO₃ solution?
Answer:
The correct answer is 1) 56 ml and 2) 0.314 M
Explanation:
1. The reaction taking place in the given case is,
HClO₃ + KOH ⇒ KClO₃ + H2O, the molarity of HClO₃ given is 0.100 M, the molarity of KOH given is 0.140 M and the volume of KOH given is 40 ml, there is a need to find the volume of HClO₃.
Therefore, the mole of HClO₃ = mole of KOH
= MHClO₃ × VHClO₃ = MKOH × VKOH
= 0.100 M × VHClO₃ = 0.140 M × 40 ml
VHClO₃ = 0.140 M × 40 ml/0.100 M
VHClO₃ = 56 ml.
2. The reaction taking place is,
2HNO₃ + Ba(OH)₂ ⇒ Ba(NO₃)₂ + 2H₂O
The volume of HNO₃ given is 25 ml, the molarity of Ba(OH)2 is 0.250 M, the volume of Ba(OH)2 is 15.7 ml, the n or the number of moles of HNO₃ is 2, and the n of Ba(OH)2 is 1, the concentration or M of HNO₃ is,
M₁V₁/n₁ = M₂V₂/n₂
M₁ × 25/ 2 = 0.25 × 15.7/1
M₁ or molarity of HNO₃ = 0.314 M
1. The volume of HClO₃ required to neutralize the KOH is 56.0 mL
2. The concentration of the original HNO₃ solution is 0.314 M
1.
First, we will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
HClO₃ + KOH → KClO₃ + H₂O
This means,
1 mole of HClO₃ is required to neutralize 1 mole of KOH
From the titration formula
[tex]\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}[/tex]
Where
[tex]C_{A}[/tex] is the concentration of acid
[tex]C_{B}[/tex] is the concentration of base
[tex]V_{A}[/tex] is the volume of acid
[tex]V_{B}[/tex] is the volume of base
[tex]n_{A}[/tex] is the mole ratio of acid
[tex]n_{B}[/tex] is the mole ratio of base
From the given information,
[tex]C_{A} = 0.100 \ M[/tex]
[tex]C_{B} = 0.140 \ M[/tex]
[tex]V_{B} = 40.0 \ mL[/tex]
From the balanced chemical equation
[tex]n_{A} = 1[/tex]
[tex]n_{B} =1[/tex]
Putting the values into the formula, we get
[tex]\frac{0.100 \times V_{A} }{0.140 \times 40.0} = \frac{1}{1}[/tex]
∴ [tex]0.100 \times V_{A} = 0.140 \times 40.0[/tex]
[tex]V_{A}=\frac{0.140\times 40.0}{0.100}[/tex]
[tex]V_{A}=\frac{5.60}{0.100}[/tex]
[tex]V_{A}=56.0 \ mL[/tex]
Hence, the volume of HClO₃ required to neutralize the KOH is 56.0 mL
2.
First, we will write the balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
2HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + 2H₂O
This means, 2 mole of HNO₃ is required to neutralize 1 mole Ba(OH)₂
From the given information,
[tex]V_{A} = 25.0\ mL[/tex]
[tex]C_{B} = 0.250 \ M[/tex]
[tex]V_{B} = 15.7 \ mL[/tex]
From the balanced chemical equation
[tex]n_{A} = 2[/tex]
[tex]n_{B} =1[/tex]
Also, Using the titration formula
[tex]\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}[/tex]
We get
[tex]\frac{C_{A} \times 25.0 }{0.250 \times 15.7} = \frac{2}{1}[/tex]
Then,
[tex]C_{A} = \frac{2\times 0.250 \times 15.7} {1 \times 25.0}[/tex]
[tex]C_{A} =\frac{7.85}{25.0}[/tex]
[tex]C_{A} =0.314 \ M[/tex]
Hence, the concentration of the original HNO₃ solution is 0.314 M
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A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup calorimeter fitted with a lid through which a thermometer passes. The acid-base reaction is as follows:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
The temperature of each solution before mixing is 22.3 °C. After mixing, the temperature of the solution mixture reaches a maximum temperature of 31.4 °C. Assume the density of the solution mixture is 1.00 g/mL, its specific heat is 4.18 J/g.°C, and no heat is lost to the surroundings. Calculate the enthalpy change, in kj, per mole of H2SO4 in the reaction.
a. +85.6 kJ/mol.
b. -85.6 kJ/mol.
c. +5.71 kJ/mol.
d. -5.71 kJ/mol.
e. -114 kJ/mol.
Answer:
THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.
Explanation:
Heat = mass * specific heat capacity * temperature rise
Total volume = 100 + 50 = 150 mL
Total mass = density * volume
Total mass = 1 * 150 mL = 150 g
So therefore, the heat evolved during the reaction is:
Heat = 150 * 4.18 * ( 31.4 - 22.3)
Heat = 150 * 4.18 * 9.1
Heat = 5705.7 J
Equation for the reaction:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water
50 mL of 1 M of H2SO4 contains
50 * 1 / 1000 mole of acid
= 0.05 mole of acid
The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:
0.05 mole of H2SO4 produces 5705.7 J of heat
1 mole of H2SO4 will produce 5705.7 / 0.05 J
= 114,114 J / mole
In kj/mole = 114 kJ/mole.
Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.
From the following balanced equation, CH4(g)+2O2(g)⟶CO2(g)+2H2O(g) how many grams of H2O can be formed when 1.25g CH4 are combined with 1.25×10^23 molecules O2? Use 6.022×10^23 mol−1 for Avogadro's number.
Answer:
2.81 g of H2O.
Explanation:
We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.
This can be obtained as follow:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.
1 mole of O2 = 16x2 = 32 g.
Thus 6.022×10²³ molecules is present in 32 g of O2,
Therefore, 1.25×10²³ molecules will be present in =
(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.
Therefore, 1.25×10²³ molecules present in 6.64 g of O2.
Next, the balanced equation for the reaction. This is given below:
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)
Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.
This can be obtained as follow:
Molar mass of CH4 = 12 + (4x1) = 16 g/mol.
Mass of CH4 from the balanced equation = 1 x 16 = 16 g
Molar mass of O2 = 16x2 = 32 g/mol.
Mass of O2 from the balanced equation = 2 x 32 = 64 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol.
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.
Next, we shall determine the limiting reactant.
This can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2.
Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.
From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.
Therefore, CH4 is the limiting reactant.
Finally, we shall determine the mass of H2O produced from the reaction.
In this case, the limiting reactant will be used because it will give the maximum yield of H2O.
The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted to produce produce 36 g if H2O.
Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.
Therefore, 2.81 g of H2O were obtained from the reaction.
The mass in grams of H₂O which can be formed when 1.25g CH₄ are combined with 1.25×10²³ molecules O₂ is 2.8 grams.
What is stoichiometry?Stoichiometry of any reaction tells about the amount of species present before and after the completion of the reaction.
Given chemical reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Moles of CH₄ will b calculate as:
n = W/M, where
W = given mass = 1.25g
M = molar mass = 16g/mol
n = 1.25/16 = 0.078 moles
Molecues of CH₄ in 0.078 moles = 0.078×6.022×10²³ = 0.46×10²³
Given molecules of O₂ = 1.25×10²³
Required molecules of CH₄ is less as compared to the molecules of O₂, so here CH₄ is the limiting reagent and formation of water is depends on it only.
From the stoichiometry of the reaction it is clear that:
1 mole of CH₄ = will produce 2 moles of H₂O
0.078 moles of CH₄ = will produce 2×0.078=0.156 moles of H₂O
Mass of H₂O will be calculated by using its moles as:
W = (0.156)(18) = 2.8g
Hence required mass of H₂O is 2.8g.
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Aqueous potassium nitrate (KNO3) and solid silver bromide are formed by the reaction of aqueous potassium bromide and aqueous silver nitrate (AgNO3). Write a balanced chemical equation for this reaction
Answer:
For the mentioned reaction, the balanced chemical equation is:
KBr (aq) + AgNO3 (s) ⇒ KNO3 (aq) + AgBr (s)
The number written in front of the ion, atoms, and molecules in a chemical reaction so that each of the elements on both the sides of reactants and products of the equation gets balanced is known as the stoichiometric coefficient.
From the mentioned balanced equation, the stoichiometric coefficient before KBr is 1, AgNO3 is 1, KNO3 is 1, as well as before AgBr is also 1. Thus, it is clear that 1 mole of potassium bromide reacts with 1 mole of silver nitrate to produce 1 mole of potassium nitrate and 1 mole of silver bromide.
Using the standard reduction potentials Ni2+(aq) + 2 e‑Ni(s) ‑0.25 volt Fe3+(aq) + e‑Fe2+(aq) +0.77 volt Calculate the value of E°cell for the cell with the following reaction. Ni2+(aq) + 2 Fe2+(aq) →Ni(s) + 2 Fe3+(aq)
Answer:
The correct answer is - 1.02 V
Explanation:
From the reduction-oxidation reaction:
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:
Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s) Eº= ‑0.25 V
Oxidation (anode) : 2 x (Fe²⁺ → Fe³⁺ + e-)(aq) Eº= -0.77 V
-------------------------------------
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):
Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V
While balancing a chemical equation, we change the _____ to balance the number of atoms on each side of the equation.
Answer:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation
Explanation:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
What is chemical equation?To summarize in chemistry terms, a chemical equation depicts the initial chemicals, or reactants, on the left-hand side and the final compounds, or products, just on right-hand side, divided by an arrow. In the chemical equation, the number of atoms in each element as well as the total charge are the same on opposite of the equation's sides.
Chemical equations are used in chemistry to depict chemical processes by writing the reactants and products in terms of their corresponding chemical formulas. While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
Therefore, while balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
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Each energy sublevel contains __________ number of electrons. For example, sublevel D can hold up to _______ electrons. A. the same, 10 B. the same, 14 C. a different, 6 D. a different, 10
Answer:
Each energy sublevel contains a different number of electrons. For example, sublevel D can contain up to 10 electrons
Explanation:
The atoms are surrounded by propellers that within each propeller there is a certain number of electrons, these electrons jump from orbit to orbit according to the amount of energy they have. The four levels that make up the electronic cloud that surrounds an atom are: s p d f.
When these electrons change orbit or level they release energy in the form of light, which is known as a photon.
A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?
Question is incomplete, the complete question is as follows:
A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?
A. Toxicity, because it can be observed by altering the state of the substance
B. Boiling point, because it can be observed by altering the state of the substance
C. Toxicity, because it can be observed by replacing the atoms of the substance with new atoms
D. Boiling point, because it can be observed by replacing the atoms of the substance with new atoms
Answer:
B.
Explanation:
A student can examine a substance without altering the bonds within the molecules by examining its boiling point.
The boiling point is the property of a substance, at which the substance changes its state, which is from solid to liquid, liquid to gas and others. So, examining the boiling point will alter the bonds within the molecules as the state of substance will change.
Hence, the correct answer is "B".
Calculate LaTeX: \DeltaΔGº for a voltaic cell with Eºcell = +0.24 V if the overall reaction involves a 3 electron reduction.
Answer:
-69 kJ
Explanation:
Step 1: Given data
Standard cell potential (E°cell): +0.24 V
Electrons involved (n): 3 mol
Step 2: Calculate the standard Gibbs free energy change (ΔG°) for the voltaic cell
We will use the following expression.
ΔG° = -n × F × E°cell
where,
F is Faraday's constant (96,485 C/mol e⁻)
ΔG° = -n × F × E°cell
ΔG° = -3 mol e⁻ × 96,485 C/mol e⁻ × 0.24 V
ΔG° = -69 kJ
Given the following equivalents, make the following conversion: 1.00 knop = ? knips
4 clips = 5 blips
1 knop = 6 bippy
3 blip = 18 pringle
1 clip = 10 knip
10 bippy = 8 pringle
Answer:
[tex]6.4knips[/tex]
Explanation:
Hello,
In this case, given the stated equivalences, we can use the following proportional factor in order to compute the required knips:
[tex]knips=1.00knop*\frac{6bippy}{1knop} *\frac{8pringle}{10bippy}* \frac{3blip}{18pringle} *\frac{4clips}{5blips} *\frac{10knip}{1clip} \\\\=6.4knips[/tex]
Regards.
1 knop=6.4 knips
First convert knop to bippy:-
[tex]1\ knop\times\frac{6\ bippy}{1\ knop} =6\ bippy[/tex]
Now, Convert 6 bippy to pringle:-
[tex]6\ bippy\times\frac{8\ pringle}{10\ bippy} =4.8\ pringle[/tex]
Now, convert 4.8 pringle to blip:-
[tex]4.8\ pringle\times\frac{3\ blip}{18\ priangle} =0.8\ blip[/tex]
Now, convert 0.8 blip to clips as follows:-
[tex]0.8\ blip\times\frac{4\ clips}{5\ blip} =0.64\ clip[/tex]
Now, convert 0.64 clip to knips:-
[tex]0.64\ clip\times\frac{10\ knip}{1\ clip} =6.4\ knip[/tex]
Hence, the following conversion is as follows:-
1.00 knop=6.4 knips
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Arrange the compounds in order of decreasing magnitude of lattice energy:
a. LiBr
b. KI
c. CaO.
Rank from largest to smallest.
Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.
Arranging the chemical compounds in order of decreasing magnitude of lattice energy, we have:
c. CaO.
a. LiBr
b. KI
Lattice energy can be defined as a measure of the energy required to dissociate one (1) mole of an ionic compound into its constituent anions and cations, in the gaseous state.
Hence, it is typically used to measure the bond strength of ionic compounds.
Generally, lattice energy is inversely proportional to the size of the ions and directly proportional to their electric charges.
Lithium bromide (LiBr) comprises the following ions:
[tex]Li^+[/tex] and [tex]Br^-[/tex]Potassium iodide (KI) comprises the following ions:
[tex]K^+[/tex] and [tex]I^-[/tex]Calcium oxide (CaO) comprises the following ions:
[tex]Ca^{2+}[/tex] and [tex]O^{2-}[/tex]From the above, we can deduce that there is an increase in the charge possessed by the ionic chemical compounds and as such this would result in an increase in the lattice energy.
In order of decreasing magnitude of lattice energy, the chemical compounds are arranged as:
I. CaO.
II. KI.
III. LiBr.
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A 1.0 L buffer solution is 0.250 M HC2H3O2 and 0.050 M LiC2H3O2. Which of the following actions will destroy the buffer?
A. adding 0.050 moles of NaOH
B. adding 0.050 moles of LiC2H3O2
C. adding 0.050 moles of HC2H3O2
D. adding 0.050 moles of HCl
E. None of the above will destroy the buffer.
Answer:
D
Explanation:
Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.
The action that destroys the buffer is option c. adding 0.050 moles of HCl.
What is acid buffer?It is a solution of a weak acid and salt.
Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.
The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this, there will be only acid in the solution.
Since
moles of HC2H3O2 = 1*0.250 = 0.250
moles of NaC2H3O2 = 1*0.050 = 0.050.
moles of HCl is added = 0.050
Now
The reaction between HCl and NaC2H3O2
[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]
Now
BCA table is
NaC2H3O2 HCl HC2H3O2
Before 0.050 0.050 0.250
Change -0.050 -0.050 +0.050
After 0 0 0.300
Now, the solution contains the acid (HC2H3O2 ) only.
Therefore addition of 0.050 moles of HCl will destroy the buffer.
Learn more about moles here: https://brainly.com/question/24817060
What can you learn about the pH of a substance with the conductivity test? hint: gives you no info on concentration.
Answer:
See explanation
Explanation:
So, I'm gonna take a shot at this one and say this:
With a strongly acidic/basic solution, you'll get a high conductivity when preforming a conductivity test.
The more acidic or basic a substance is, the higher the electrical conductivity.
Based on how high or low the conductivity is, it will give you an idea of the substance's pH.
Hope that made since or gave you an idea of what you're looking for. Good luck :)
Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
Find the volume if the density is 2.6 g/mL and the mass is 9.7 g.(5 pts)
Find the mass if the density is 1.6 g/cm3 and the volume is 4.1 cm3 (5 pts)
Find the density if the initial volume of water is 12.8 mL, the final volume is 24.6 mL and the mass of the object is 4.3 g. Make a drawing to show the water displacement using a graduated cylinder. (gdoc, gdraw)
Answer:
[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]
Explanation:
1. Density from mass and volume
[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]
2. Volume from density and mass
[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]
3. Mass from density and volume
[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]
4. Density by displacement
Volume of water + object = 24.6 mL
Volume of water = 12.8 mL
Volume of object = 11.8 mL
[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]
Your drawing showing water displacement using a graduated cylinder should resemble the figure below.
Which part of an atom is mostly empty space?
A. nucleus
B. proton cloud
C. electron cloud
D. neutron
Answer:
C. Electron cloud
the electron is around 1/2000 times the size of the proton.
If you imagine the proton a a marble in the middle of a football field, the electrons will revolve around the last row
A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g). Find the ratio of the rms speed of the argon molecules to that of the hydrogen. Assume hydrogen molecule has only translational degree of freedom.
Answer:
Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1
Explanation:
The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:
Vrms = [tex]\sqrt{3RT/M}[/tex]
where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol
For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R
Vrms = √(3 * R *4T)/0.04 = √300RT
For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R
Vrms = √(3 * R *T)/0.001 = √3000RT
Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316
Ratio of Vrms of argon to that of hydrogen = 0.316 : 1
Provide the name(s) for the tertiary alcohol(s) with the chemical formula C6H14O that have a 4-carbon chain. Although stereochemistry may be implied in the question, DO NOT consider stereochemistry in your name. Alcohol #1______ Alcohol #2: ______Alcohol #3______
Answer:
Explanation:
A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.
Based on the question, the only tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain is
2-hydroxy-2,3-dimethylbutane
H OH H H
| | | |
H - C - C - C - C - H
| | | |
H CH₃ CH₃ H
From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain
Which of the following is a salt that could be generated by combining a weak acid and a weak base? Select the correct answer below:
a) NaCl
b) Na2SO4
c) NH4NO3
d) NH4F
Answer:
d) NH4F
Explanation:
Hello,
In this case, the base resulting from mixing a weak acid and a weak base is d) NH4F since ammonium hydroxide is a wear base and hydrofluoric acid is a weak acid.
Ammonium hydroxide is a weak base since it is not completely ionized in ammonium and hydroxyl ions:
[tex]NH_4OH\rightarrow NH_4^++OH^-[/tex]
Moreover, hydrofluoric acid is a weak acid since it is not completely ionized in hydrogen and fluoride ions:
[tex]HF\rightleftharpoons H^++F^-[/tex]
For the both of the substances, the limit is established by the basic and the acid dissociation constant respectively.
Regards.
