The cell potential for the electrochemical cell has been 1.40 V.
The standard reaction for the cell will be:
[tex]\rm 3\;Cl_2\;+\;2\;Fe\;\rightarrow\;6\;Cl^-\;+\;2\;Fe^3^+[/tex]
The half-reaction of the cells has been:
[tex]\rm Fe^3^+\;+\;3\;e^-\;\rightarrow\;Fe[/tex]
The potential for this reduction has been -0.04 V.
[tex]\rm Cl_2\;+\;2\;e^-\;\rightarrow\;2\;Cl^-[/tex]
The potential for the reduction has been 1.36 V.
The cell potential has been: Potential of reduction - Potential of oxidation
Cell potential = 1.36 - (-0.04) V
Cell potential = 1.40 V.
The cell potential for the electrochemical cell has been 1.40 V.
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What is the oxidizing agent in the redox reaction represented by the following cell notation? Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
Answer:
Silver.
Explanation:
Hello,
In this case, for the redox reaction:
[tex]Ni^0(s)+Ag^+(aq)\rightarrow Ni^{2+}+Ag^0(s)[/tex]
We can see the nickel is being oxidized as its oxidation state increases from 0 to 2+ whereas the oxidation state of silver decreases from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.
Best regards.
The oxidizing agent in the redox reaction represented by the following cell notation is Silver.
Calculation of the oxidizing agent:The redox reaction is
Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
here the nickel is being oxidized since its oxidation state rises from 0 to 2+ while on the other hand, the oxidation state of silver is reduced from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.
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In the experiment students will create solutions with different ratios of ethanol and water. What is the mole fraction of ethanol when 10.00 mL of pure ethanol is combined with 2.00 mL of water
Answer:
[tex]x_{et}=0.6068[/tex]
Explanation:
Hello,
In this case, since the mole fraction of a compound, in this case ethanol, in a binary mixture, in this constituted by both water and ethanol, is mathematically defined as follows:
[tex]x_{et}=\frac{n_{et}}{n_{et}+n_{w}}[/tex]
Whereas [tex]n[/tex] accounts for the moles in the solution for each species, we must first compute the moles of both ethanol (density: 0.789 g/mL and molar mass: 46.07 g/mol) and water (density: 1g/mL and molar mass: 18.02 g/mol)
[tex]n_{et}=10.00mL\ et*\frac{0.789g\ et}{mL\ et} *\frac{1mol\ et}{46.07g\ et}=0.1713mol\ et\\ \\n_w=2.00mL\ w*\frac{1g\ w}{mL\ w} *\frac{1mol\ w}{18.02g\ w}=0.1110mol\ w[/tex]
Therefore, the mole fraction turns out:
[tex]x_{et}=\frac{0.1713mol}{0.1713mol+0.1110mol}\\\\x_{et}=0.6068[/tex]
Best regards.
write the balanced nuclear equation for the radioactive decay of radium-226 to give radon-222, and determine the type of decay
Answer:
226Ra88→222Rn86+4He2
Explanation:
An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.
During α-decay(alpha decay), an atomic nucleus emits an alpha particle.
1. For the following reaction, 4.86 g of magnesium nitride are mixed with excess water. The reaction yields 7.18 g of magnesium hydroxide.
magnesium nitride(s) + water(1) –> magnesium hydroxide (aq) + ammonia (aq)
What is the ideal yield of magnesium hydroxide?
What is the percent yield for this reaction?
2. For the following reaction, 6.41 g of hydrogen gas are mixed with excess nitrogen gas. The reaction yields 26.2 g of ammonia.
nitrogen(g) + hydrogen(g) –> ammonia(g)
What is the ideal yield of ammonia?
What is the percent yield for this reaction?
3. For the following reaction, 3.79 g of water are mixed with excess chlorine gas. The reaction yields 8.70 g of hydrochloric acid.
chlorine(g) + water(1) –> hydrochloric acid(aq) + chloric acid (HCIO3)(aq)
What is the ideal yield of hydrochloric acid?
What is the percent yield for this reaction?
Answer:
See explanation
Explanation:
1)
Mg3N2(s) + 6H2O(l) ------------> 3Mg(OH)2 + 2NH3(g)
Number of moles of magnesium nitride= mass/molar mass= 4.86g/100.9494 g/mol = 0.048 moles
1 mole of magnesium nitride yields 3 moles of magnesium hydroxide
0.048 moles of magnesium nitride yields 0.048 moles × 3= 0.144 moles of magnesium hydroxide
Theoretical yield of magnesium hydroxide = 0.144 moles × 58.3197 g/mol = 8.398 g
Percent yield= actual yield/ theoretical yield × 100
Percent yield= 7.18/8.398 × 100/1 = 85.5%
2)
N2(g) + 3H2(g) -------> 2NH3(g)
Number of moles of hydrogen gas = mass/ molar mass = 6.41g/ 2gmol-1 = 3.205 moles of hydrogen gas.
From the balanced reaction equation;
3 moles of hydrogen gas yields 2 moles of ammonia
3.205 moles of hydrogen gas yields 3.205 × 2/3 = 2.1367 moles of ammonia
Theoretical yield of ammonia = 2.1367 moles × 17 gmol-1 = 36.3 g
Percent yield = actual yield/ theoretical yield ×100
Percent yield = 26.2/36.3 ×100
Percent yield = 72.2%
3)
3Cl2(g) + 3H2O(l) ------> HOCl3(aq) + 5HCl(aq)
Number of moles of water= mass/ molar mass = 3.79g/18 gmol-1 = 0.21 moles
Since
3 mole of water yields 5 mole of HCl
0.21 moles of water yields 0.21 × 5/3 = 0.35 moles of HCl
Theoretical yield of HCl = 0.35 moles × 36.5 gmol-1 = 12.775 g
Percent yield = actual yield/ theoretical yield × 100/1
Percent yield = 8.70/12.775 ×100
Percent yield = 68.1%
2. Find the two generic molecules from Part 1 that are made of 3 atoms. a. Compare and contrast these two molecules by listing two similarities and two differences.
Answer:
hello the molecules are missing from your question below are the Generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]
answer : It can be determined that both generic molecules are polar
It can be determined that both generic molecules have similar molecular shape
They have different Geometry
They differ in bond angles as well
Explanation:
The two generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]
comparing(similarities) these two generic molecules
It can be determined that both generic molecules are polar
It can be determined that both generic molecules have similar molecular shape
differences between the generic molecules
They have different Geometry
They differ in bond angles as well
For each of the following systems at equilibrium, predict whether the reaction will shift to the right, left, or not be affected by a decrease in the reaction container volume.
A) PCl3(g) + Cl2(g) ⇌ PCl5(g) 1. no shift
2. right shift
3. left shift B) 2 NO(g) ⇌ N2(g) + O2(g)
1. no shift 2. right shift3. left shift C) 2 NO2(g) ⇌ N2O4(g)
1. no shift 2. right shift3. left shift
Answer:
A - right shift
B - no shift
C - right shift
Explanation:
According to Le Chatelier's principle, when a reaction is in equilibrium and one of the factors affecting the rate of reaction is introduced, the equilibrium will shift so as to annul the effects of the constraint.
In this case, decreasing the volume of the reaction's container is equivalent to increasing the pressure of the reaction. When pressure is increased, the reaction will shift towards the side with a lower number of moles.
In A, the total number of moles on the left-hand side of the reaction is two while it is one on the right-hand side. An increase in pressure will, therefore, see the equilibrium shifting to the right-hand side.
In B, the total number of moles on both the right and the left-hand side is two each. An increase in the pressure will have no effect on the equilibrium.
In C, the total number of moles on the left-hand side is two while it is one on the right-hand side. Hence, an increase in the pressure of the reaction will cause a shift in the equilibrium to the right.
The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes has elapsed? Report your answer to 1 decimal place.
Answer:
12.5g
Explanation:
Half life = 2.4 Minutes.
The half life of a compound is the time it takes to decay to half of it's original concentration or mass.
Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)
Initial mass = 100g
First half life;
100g --> 50g
Second half life;
50g --> 25g
Third half life;
25g --> 12.5g
The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 2.4 mins
Time (t) = 7.2 mins
Number of half-lives (n) =?[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]
n = 3Thus, 3 half-lives has elapsed.
Finally, we shall the amount remaining. This can be obtained as follow:Original amount (N₀) = 100 g
Number of half-lives (n) = 3
Amount remaining (N) =?[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]
N = 12.5 gThus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g
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Which response includes all the following processes that are accompanied by an increase in entropy? 1) 2SO 2(g) + O 2(g) → SO 3(g) 2) H 2O(l) → H 2O(s) 3) Br 2(l) → Br 2(g) 4) H 2O 2(l) → H 2O(l) + 1/ 2O 2(g)
Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]
3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.
2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]
1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.
3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]
1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.
4) [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]
1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.
Plz help!!!! Solve this by using factor labeling
Answer:
the answer is 2,000 nickels.
Explanation:
we multiplied 100 by 100, because there are 100 cents in a dollar, and we divided 10,000 by 5, because there are 5 cents in a nickel.
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K sp (MgF 2) = 6.9 × 10 –9]
Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]
And the undergoing chemical reaction:
[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]
We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:
[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]
Next, the moles of magnesium chloride consumed by the sodium fluoride:
[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]
Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:
[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]
[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]
Thereby, the reaction quotient is:
[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.
Assume that 33.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of the monoprotic strong acid HX.How many moles of have been added at the equivalence point?n = ? mol
Which of the following sets of quantum numbers (n, l, ml, ms) refers to a 3d orbital? Question 5 options: 2 0 0 – g 5 4 1 – 4 2 –2 + 4 3 1 – 3 2 1 –
3 2 1 is the set of quantum numbers.
What are Quantum Numbers?The set of numbers used to describe the position and energy of the electron in an atom is called quantum numbers. There are four quantum numbers, namely, principal, azimuthal, magnetic, and spin quantum numbers.
What is the rule of quantum numbers?The rules for quantum numbers are: (n) can be any positive, nonzero integral value. (l) can be zero or any positive integer but not larger than (n-1). l = 0, 1, 2, 3, 4, …. (n-1) (ml) values follow the equation.
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clacium hydroxide is slightly soluable in water about 1 gram will dissolve in 1 liter what are the spectator ions in the reaction ions in the reaction of such a dilute solution of calcium hydroxide with hydrochloric acid
Answer:
Ca²⁺ and Cl⁻
Explanation:
In a chemical reaction, spectator ions are ions that are not involved in the reaction, that means are the same before and after the reaction.
In water, calcium hydroxide, Ca(OH)₂ is dissociated in Ca²⁺ and OH⁻. Also, hydrochloric acid, HCl, dissociates in H⁺ and Cl⁻. The reaction is:
Ca²⁺ + 2OH⁻ + 2H⁺ + 2Cl⁻ → 2H₂O + Ca²⁺ + 2Cl⁻
The ions that react are H⁺ and OH⁻ (Acid and base producing water)
And the ions that are not reacting, spectator ions, are:
Ca²⁺ and Cl⁻which law states that the pressure and absolue tempeture of a fixed quantity of gas are directly proportional under constant volume conditions?
Answer:
Gay lussacs law
Explanation:
How many liters of CH₃OH gas are formed when 3.20 L of H₂ gas are completely reacted at STP according to the following chemical reaction? Remember 1 mol of an ideal gas has a volume of 22.4 L at STP.CO(g)+ H2(g) → CH3OH
Answer:
The correct answer is 1.60 Liters.
Explanation:
The given reaction:
CO (g) + H₂(g) ⇔ CH₃OH (g)
Based on the given reaction, two moles of H₂ reacts with one mole of CO and produce one mole of CH₃OH.
It is mentioned that 3.20 L of H₂ is reacted, therefore, there is a need to convert it into moles.
As 22.4 L at standard temperature and pressure is equivalent to 1 mole.
Therefore, 1 L at STP will be, 1/22.4 mole
Now 3.20 L at STP will be,
= 1/22.4 × 3.20
= 0.1428 mole
And as mentioned in the reaction that 2 moles of H₂ gives 1 mole of CH₃OH, therefore, 1 mole of H₂ will give 1/2 mole of CH₃OH
Now, 0.1428 mole of H₂ will give,
= 0.1428/2 = 0.071 mole of CH₃OH
= 0.071 × 22.4 = 1.60 L
The volume, in liters, of CH₃OH gas formed is 1.60 L
From the question,
We are to determine the volume of CH₃OH formed
The given chemical equation for the reaction is
CO(g)+ H₂(g) → CH₃OH
The balanced chemical equation for the reaction is
CO(g)+ 2H₂(g) → CH₃OH
This means
1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH
Now, we will determine the number of moles of H₂ present in the 3.20 L H₂ at STP
1 mol of an ideal gas has a volume of 22.4 L at STP
Then,
x mole of the H₂ gas will have a volume of 3.20 L at STP
x = [tex]\frac{3.20 \times 1}{22.4}[/tex]
x = 0.142857 mole
∴ The number of mole of H₂ present is 0.142857 mole
Since
2 moles of H₂ reacts to produce 1 mole of CH₃OH
Then,
0.142857 mole of H₂ will react to produce [tex]\frac{0.142857}{2}[/tex] mole of CH₃OH
[tex]\frac{0.142857}{2} = 0.0714285[/tex]
∴ The number of moles of CH₃OH produced = 0.0714285 mole
Now, for the volume of CH₃OH formed
Since
1 mol of an ideal gas has a volume of 22.4 L at STP
Then,
0.0714285 mol of CH₃OH will have a volume of 22.4 × 0.0714285 at STP
22.4 × 0.0714285 = 1.5999984 L ≅ 1.60 L
Hence, the volume of CH₃OH gas formed is 1.60 L
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What would happen to the rate of a reaction with rate law rate = k [NO]2[Hz] if
the concentration of NO were doubled?
Answer:
The rate would have doubled
Explanation:
g When considering the effects of temperature on spontaneity, if both ΔH and ΔS are positive, _______. Select the correct answer below: the process is spontaneous at all temperatures
Explanation:
The spontaneity of a system is deduced by the sign of the gibbs free energy value. If it is negative, it means the process / reaction is spontaneous however a positive value indicates the such process is not spontaneous.
Gibbs free energy, enthalpy and entropy are related by the following equation;
ΔG = ΔH - TΔS
A positive value of enthalpy, H and entropy, S means that G would always be a negative value at all temperatures.
A sample of a hydrocarbon is found to contain 7.99g carbon and 2.01g hydrogen. What is the empirical formula for this compound
Answer:
The empirical formulae for the compound is CH3.
A baseball has a mass of 0.145 kilograms. If acceration due to gravity is 9.8m/s,what is the weight of the baseball in newtons?
Answer:
I hope it works
Explanation:
As we know that
w=m*g
given m=0.145 , g=9.8
hence we get
w= (9.8)*(0.145)
w=1.421 m/sec 2
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To find the pH of a solution of NH4Br directly, one would need to use:__________
Select the correct answer below:
a) the Kb of NH3 to find the hydroxide concentration
b) the Ka of NH+4 to find the hydronium concentration
c) the Kb of NH3 to find the hydronium concentration
d) the Ka of NH+4 to find the hydroxide concentration
Answer:
b) the Ka of NH₄⁺ to find the hydronium concentration
Explanation:
The equilbrium of NH₄⁺ (The conjugate acid of NH₃, a weak base), is:
NH₄⁺ ⇄ NH₃ + H⁺
Where Ka of the conjugate acid is:
Ka = [NH₃] [H⁺] / [NH₄⁺]
Thus, if you know Ka of NH₄⁺ and its molar concentration you can calculate [H⁺], the hydronium concentration, to find pH (Because pH = -log [H⁺])
Thus, right option is:
b) the Ka of NH₄⁺ to find the hydronium concentrationThe combination of a carbonyl group and a hydroxyl group on the same carbon atom is called a ________ group.
a. carbamate group
b. carbonate
c. carboxlate
d. carboxyl
Answer:
d. carboxyl
Explanation:
The presence of carbonyl group (>C=O)) and a hydroxyl group ( (−OH) on the same carbon atom is called a "carboxyl" group. A carboxyl group is represented as COOH and acts as the functional group part of carboxylic acids.
For example:
Formic acid or Methanoic acid (H-COOH) Butanoic acid (C3H7-COOH)Hence, the correct option is "d. carboxyl ".
Hydrazine, , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine used as a fuel for rockets: How many moles of each of the gaseous products are produced when 20.1 g of pure hydrazine is ignited in the presence of 20.1 g of pure oxygen
Answer:
[tex]1.25~mol~H_2O[/tex] and [tex]0.627~mol~N_2[/tex]
Explanation:
Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ([tex]N_2H_4[/tex]) and oxygen ([tex]O_2[/tex]). So, we can start with the reaction between these compounds:
[tex]N_2H_4~+~O_2~->~N_2~+~H_2O[/tex]
Now we can balance the reaction:
[tex]N_2H_4~+~O_2~->~N_2~+~2H_2O[/tex]
In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for [tex]N_2H_4[/tex] and 31.99 g/mol for [tex]O_2[/tex]):
[tex]20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4[/tex]
[tex]20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2[/tex]
In the balanced reaction we have 1 mol for each reagent (the numbers in front of [tex]O_2[/tex] and [tex]N_2H_4[/tex] are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is [tex]N_2H_4[/tex].
With this in mind, we can calculate the number of moles for each product. In the case of [tex]N_2[/tex] we have a 1:1 molar ratio (1 mol of [tex]N_2[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:
[tex]0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2[/tex]
We can follow the same logic for the other compound. In the case of [tex]H_2O[/tex] we have a 1:2 molar ratio (2 mol of [tex]H_2O[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:
[tex]0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O[/tex]
I hope it helps!
If the Ksp for Li3PO4 is 5.9×10−17, and the lithium ion concentration in solution is 0.0020 M, what does the phosphate concentration need to be for a precipitate to occur?
Answer:
7.4 × 10⁻⁹ M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷
Concentration of lithium ion: 0.0020 M
Step 2: Write the reaction for the solution of Li₃PO₄
Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
Step 3: Calculate the phosphate concentration required for a precipitate to occur
The solubility product constant is:
Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³
[PO₄³⁻] = 7.4 × 10⁻⁹ M
Automotive air bags inflate when sodium azide decomposes explosively to its constituent elements: 2NaN3 (s) → 2Na (s) + 3N2 (g) How many grams of sodium azide are required to produce 30.5 g of nitroge
Answer:
NaN3 = 47.2 g
Explanation:
Given:
2 NaN3 ⇒ 2 Na + 3 N2
Find:
Amount of NaN3
Computation:
N2 moles = Product of N2 / molar mass of N2
N2 moles =30.5/28
N2 moles = 1.0893
2NaN3 makes 3(N2 )
So,
NaN3 moles = (2/3) moles of N2
NaN3 moles = ( 2/3) × 1.0893
NaN3 moles = = 0.7262
NaN3 mass = 0.7262 x 65
NaN3 = 47.2 g
Answer:
NaN3 = 47.2 g
Explanation:
Given:
2 NaN3 ⇒ 2 Na + 3 N2
Find:
Amount of NaN3
Computation:
N2 moles = Product of N2 / molar mass of N2
N2 moles =30.5/28
N2 moles = 1.0893
2NaN3 makes 3(N2 )
So,
NaN3 moles = (2/3) moles of N2
NaN3 moles = ( 2/3) × 1.0893
NaN3 moles = = 0.7262
NaN3 mass = 0.7262 x 65
NaN3 = 47.2 g
Explanation:
The ph of the test tube can be calculated by knowing the concentration of hydroxide ions in solution. the ph = 14 + log 10 oh- for example the 0.1m stock of naoh has a ph = 14+ log 10 0.1= 13. using the dilution 5 ml 0.1 naoh 5ml h20 what is the ph of tube 1.
Answer:
Volume of the calcium hydroxide solution used is 0.0235 mL.
Explanation:
Moles of KHP =
According to reaction, 2 moles of KHP with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;
of calcium hydroxide
Molarity of the calcium hydroxide solution = 0.703 M
Volume of calcium hydroxide solution = V
Volume of the calcium hydroxide solution used is 0.0235 mL.
what is the molality of a solution
Answer: The number of moles of a solute per kilogram of solvent
Explanation:
Consider the reaction for the dissolution of solid magnesium hydroxide.
Mg(OH)2(s)g2 (a) +2OH (ag)
If the concentration of hydroxide ion in a saturated solution of magnesium hydroxide is 2.24 x 104 M.
what is the molar solubility of magnesium hydroxide? Report your answer in scientific notation with three significant figures.
Answer:
Molar solubility is 1.12x10⁻⁴M
Explanation:
The dissolution of magnesium hydroxide is:
Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻
The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:
Mg(OH)₂(s) ⇄ X + 2X
Where X is solubility.
If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:
2X = 2.24x10⁻⁴M
X = 2.24x10⁻⁴M/2
X =1.12x10⁻⁴M
Molar solubility is 1.12x10⁻⁴M
PV = nRT. If P = 1 atm, V = 5.0 liter, R = 0.0821 L.atm/mol.K, and T = 293 K; what is the value of n?
Answer:
n = 0.207 mole
Explanation:
We have,
P = 1 atm
V = 5 liter
R = 0.0821 L.atm/mol.K
T = 293 K
We need to find the value of n. The relation is as follows :
PV = nRT
Solving for n,
[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}[/tex]
So, the value of n is 0.207 mol.
According to the following reaction, how many moles of ammonia will be formed upon the complete reaction of 31.2 grams of nitrogen gas with excess hydrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answer:
4.46 mol of NH3
Explanation:
The equation of he reaction is given as;
2N + 3H2 --> 2NH3
From the stochiometry of the reaction, 1 mol of Nitrogen produces 2 mol of Ammonia.
Mass of Nitrogen = 31.2g
Molar mass of Nitrogen = 14g/mol
Number of moles = Mass / Molar mass = 31.2 / 14 = 2.23 mol
Since 1 mol of N = 2 mol of NH3;
2.23 mol of N2 would produce x
x = 2.23 * 2 = 4.46 mol of NH3
If the rate of formation (also called rate of production) of compound C is 2M/s in the reaction A --->2C, what is the rate of consumption of A
Answer:
[tex]r_A=-1\frac{M}{s}[/tex]
Explanation:
Hello,
In this case, given the rate of production of C, we can compute the rate of consumption of A by using the rate relationships which include the stoichiometric coefficients at the denominators (-1 for A and 2 for C) as follows:
[tex]\frac{1}{-1} r_A=\frac{1}{2}r_C[/tex]
In such a way, solving the rate of consumption of A, we obtain:
[tex]r_A=-\frac{1}{2} r_C=-\frac{1}{2}*2\frac{M}{s}\\ \\r_A=-1\frac{M}{s}[/tex]
Clearly, such rate is negative which account for consumption process.
Regards.
