An aqueous solution of potassium bromide, KBr, contains 4.34 grams of potassium bromide and 17.4 grams of water. The percentage by mass of potassium bromide in the solution is 20 %.

Answers

Answer 1

Answer:

True

Explanation:

The percentage by mass of a substance in a solution can be calculated by dividing the mass of the substance dissolved in the solution by the total mass of the solution. This can be expressed mathematically as:

Percentage by mass = mass of substance in solution/mass of solution x 100

In this case;

mass of KBr = 4.34 grams

mass of water = 17.4 grams

mass of solution = mass of KBr + mass of water = 4.34 + 17.4 = 21.74

Percentage by mass of KBr = 4.34/21.74 x 100

                                              = 19.96 %

19.96 is approximately 20%.

Hence, the statement is true.


Related Questions

Which statements about spontaneous processes are true? Select all that apply: A spontaneous process is one that occurs very quickly.

Answers

Answer: Here are the complete options.

A spontaneous process is one that occurs very quickly. A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium. A spontaneous process is one that occurs without continuous input of energy from outside the system. A process is spontaneous if it must be continuously forced or driven.

The correct option is

A spontaneous process is one that occurs without continuous input of energy from outside the system.

A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium

Explanation:

spontaneous process is one that occurs without continuous input of energy from outside the system and occur on its own because spontaneous processes are thermodynamically favorable characterized by a decrease in the system's free energy, they do not need to be driven by an outside source of energy. Which means that the initial energy is higher than the final energy.

A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium which will result to The sign of ΔG will change from positive to negative (or vice versa) where T = ΔH/ΔS. In cases where ΔG is: negative

Write a net ionic equation for the reaction that occurs when aqueous solutions of hydrofluoric acid and sodium hydroxide are combined. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If needed, use H for the hydronium ion.)

Answers

Answer:

The net ionic reaction is : H⁺ (aq) + OH⁻ (aq) ---> H₂O (l)

Explanation:

The reaction between aqueous solutions of hydrofluoric acid and sodium hydroxide is an example of a neutralization reaction.

A neutralization reaction is a reaction between and acid and an abase to produce salt and water only.

Hydrofluoric acid is the acid while sodium hydroxide is the base. during the reaction the hydrofluoric acid will produce hydrogen and fluoride ions, while sodium hydroxide will produce hydroxide and sodium ions. The hydroxide and hydrogen ions will combine to produce water while the sodium and fluoride ions remain in solution as ions.

The equation of the reaction is as follows:

H⁺F⁻ (aq) + Na⁺OH⁻ (aq)  ----> Na⁺F⁻ (aq) + H₂O (l)

Since the sodium and fluoride ions appear on both sides of the equation, they are known as spectator ions and are cancelled out to give the net ionic equation.

The net ionic reaction is : H⁺ (aq) + OH⁻ (aq) ---> H₂O (l)

If the theoretical yield of a reaction is 332.5 g and the percent yield for the reaction is 38 percent, what's the actual yield of product in grams? \

A. 8.74 g
B. 12616 g
C. 116.3 g
D. 126.4 g

Answers

Answer: D - 126.4g

Explanation:  

% Yield = Actual Yield/Theoretical Yield

38% = Actual Yield/332.5

38/100 = Actual Yield/332.5

(.38)(332.5) = 126.35 g = 126.4 g Actual Yield

Answer:

is D. the correct answer

Explanation:

I'm not sure if it is. Please let me know if I'm mistaking.

Consider the following chemical equation: NH4NO3(s)⟶NH+4(aq)+NO−3(aq) What is the standard change in free energy in kJmol at 298.15K? The heat of formation data are as follows: ΔH∘f,NH4NO3(s)=-365.6kJmolΔH∘f,NH+4(aq)=-132.5kJmolΔH∘f,NO−3(aq)=-205.0kJmol The standard entropy data are as follows: S∘NH4NO3(s)=151.1Jmol KS∘NH+4(aq)=113.4Jmol KS∘NO−3(aq)=146.4Jmol K Your answer should include two significant figures.

Answers

Answer:

[tex]\Delta _rG=-4.3\frac{kJ}{mol}[/tex]

Explanation:

Hello,

In this case, for the given dissociation reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):

[tex]\Delta _rH=\Delta _fH_{NH^{4+}}+\Delta _fH_{NO_3^-}-\Delta _fH_{NH_4NO_3}\\\\\Delta _rH=-132.5+(-205.0)-(-365.6)=28.1kJ/mol[/tex]

Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):

[tex]\Delta _rS=S_{NH^{4+}}+S_{NO_3^-}-S_{NH_4NO_3}\\\\\Delta _rS=113.4+146.4-151.1=108.7J/mol*K[/tex]

Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the given temperature (298.15 K), we finally obtain (two significant figures):

[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\\Delta _rG=28.1kJ/mol-(298.15 K)(108.7\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\\Delta _rG=-4.3\frac{kJ}{mol}[/tex]

Best regards.

The frequency of a signal is found to be 6389 with an uncertainty of 436 Hz. To the correct number of significant digits, it should be reported as:

Answers

Answer:

i hope it work

Explanation:

as

accurate reading range = reading ± uncertainty

so you have to say about accurate reading that its,lies in range

=(6389-436) →(6389+436)

=5953→6825

and  the correct number of significant would be 3

How are pH and pOH ?

A. pH = 14 + pOH
B. pOH = 14 - pH
C. pOH = 14 + pH
D. pH = 14 - pOH

Answers

Answer:

B. pOH = 14 - pH and D. pH = 14 - pOH.

Explanation:

Hello,

In this case, we must remember that pH and pOH are referred to a measure of acidity and basicity  respectively, since pH accounts for the concentration of H⁺ and pOH for the concentration of OH⁻ in a solution. In such a way, since the maximum scale is 14, we say that the addition between the pH and pOH must be 14:

[tex]pH+pOH=14[/tex]

Therefore, the correct answers are B. pOH = 14 - pH and D. pH = 14 - pOH since the both of them are derived from the previous definition.

Best regards.

Answer:

D: by subtracting the pOH from 14.

Explanation:

How many moles of NaOH is needed to neutralize 45.0 ml of 0.30M H2SeO4? Question 2 options: A) 0.00675 B) 27.0 C) 0.027 D) 0.0135

Answers

Answer:

C) 0.027

Explanation:

In this case we can start with the reaction between [tex]NaOH[/tex] and [tex]H_2SeO_4[/tex], so:

[tex]H_2SeO_4~+~NaOH~->~Na_2SeO_4~+~H_2O[/tex]

We have an acid ([tex]H_2SeO_4[/tex]) and a base ([tex]NaOH[/tex]), therefore we will have an acid-base reaction in which a salt is produced ([tex]Na_2SeO_4[/tex]) and water ([tex]H_2O[/tex]).

Now we can balance the reaction:

[tex]H_2SeO_4~+~2NaOH~->~Na_2SeO_4~+~2H_2O[/tex]

If we have the volume (45 mL= 0.045 L) and the concentration (0.3 M) of the acid we can calculate the moles using the molarity equation:

[tex]M=\frac{mol}{L}[/tex]

[tex]0.3~M~=~\frac{mol}{0.045~L}[/tex]

[tex]mol=0.3~M*0.045~L=0.0135~mol~of~H_2SeO_4[/tex]

In the balanced reaction, we have a 2:1 molar ratio between the acid and the base (for each mol of [tex]H_2SeO_4[/tex] 2 moles of [tex]NaOH[/tex] are consumed), with this in mind we can calculate the moles of NaOH:

[tex]0.0135~mol~of~H_2SeO_4\frac{2~mol~NaOH}{1~mol~of~H_2SeO_4}=0.027~mol~NaOH[/tex]

I hope it helps!

Question 14 (5 points)
What's the acid ionization constant for an acid with a pH of 2.11 and an equilibrium
concentration of 0.30 M?
O A) 4.87x10-8
B) 1.99x10-6
C) 3.32x10-4
OD) 2.01x10-4

Answers

Answer:

D) 2.01 x 10⁻⁴ .

Explanation:

pH = 2.11

[ H⁺ ] = [tex]10^{-2.11}[/tex]  

Let the acid be HA

It will ionise as follows .

                                        HA       ⇄       H⁺       +        A⁻

in equilibrium                 .30               [tex]10^{-2.11}[/tex]          [tex]10^{-2.11}[/tex]         

Acid ionisation constant Ka  =   [tex]\frac{(10^{-2.11})^2}{0.3}[/tex]

= 2 x 10⁻⁴                

Answer:

D) 2.01 x 10⁻⁴ is correct!

Explanation:

I got it in class!

Hope this Helps!! :))

What happens to the rate of dissolution as the temperature is increased in a gas solution?

A.
The rate stays the same.
B.
The rate decreases.
C.
The rate increases.
D.
There is no way to tell.

Answers

Answer:

The rate decreases

Explanation:

When we dissolve a gas in a water, the process is exothermic. This implies that heat is evolved upon dissolution of a gas in water.

Recall from Le Chateliers principle that for exothermic reactions, an increase in temperature favours the reverse reaction. The implication of these is that when the temperature of the gas is increased, less gas will dissolve in water.

Hence increase in temperature decreases the rate of solubility of a gas in water.

Answer:

B.

The rate decreases.

Explanation:

The decomposition of nitramide in aqueous solution at 25 °C NH2NO2(aq)N2O(g) + H2O(l) is first order in NH2NO2 with a rate constant of 4.70×10-5 s-1. If an experiment is performed in which the initial concentration of NH2NO2 is 0.384 M, what is the concentration of NH2NO2 after 31642.0 s have passed? M

Answers

Answer:

[tex][NH_2NO_2]=0.0868M[/tex]

Explanation:

Hello,

In this case, for the given chemical reaction, the first-order rate law is:

[tex]r=\frac{d[NH_2NO_2]}{dt} =-k[NH_2NO_2][/tex]

Which integrated is:

[tex][NH_2NO_2]=[NH_2NO_2]_0exp(-kt)[/tex]

Thus, the concentration after 31642.0 s for a 0.384-M solution is:

[tex][NH_2NO_2]=0.384M*exp(-4.70x10^{-5}s^{-1}*31642.0s)\\[/tex]

[tex][NH_2NO_2]=0.0868M[/tex]

Best regards.

Answer:

[A] = 0.0868 M

Explanation:

Rate constant = 4.70×10-5 s-1

First order reaction

Initial concentration, [A]o = 0.384 M

Final concentration, [A] = ?

Time, t = 31642.0 s

All these variables are related by the following equation;

[A] = [A]o e^(-kt)

[A] = 0.384  e^(-4.70×10-5 x  31642.0)

[A] = 0.384 e^(-1.4872)

[A] = 0.384 * 0.2260

[A] = 0.0868 M

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution before the addition of any HNO3. The Kb of NH3 is 1.8 × 10-5.

Answers

Answer:

[tex]pH=11.12[/tex]

Explanation:

Hello,

In this case, ammonia dissociation is:

[tex]NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)[/tex]

So the equilibrium expression:

[tex]Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]

That in terms of the reaction extent and the initial concentration of ammonia is written as:

[tex]1.8x10^{-5}=\frac{x*x}{0.10M-x}[/tex]

Thus, solving by using solver or quadratic equation we find:

[tex]x=0.00133M[/tex]

Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:

[tex]pOH=-log([OH^-])=-log(0.00133)=2.88[/tex]

And the pH from the pOH is:

[tex]pH=14-pOH=14-2.88\\\\pH=11.12[/tex]

Best regards.

Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.

Answers

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

place the following substances in Order of decreasing boiling point H20 N2 CO

Answers

Answer:

-195.8º < -191.5º < 100º

Explanation:

Water, or H20, starts boiling at 100ºC.

Nitrogen, or N2, starts boiling at -195.8ºC.

Carbon monoxide, or C0, starts boiling at -191.5ºC.

When we place these in order from decreasing boiling point:

-195.8º goes first, then -191.5º, and 100º goes last.

Answer:

therefore, N2, CO, H20

Decreasing boiling point

Explanation:

the bond existing in H2O is hydrogen bond

bond existing in N2 is covalent bond, force existing is dipole-dipole-interaction

bond existing in CO is covalent bond , force existing between is induced -dipole- induced dipole-interaction

hydrogen bond is the strongest , followed by dipole-dipole-interaction and induced -dipole- induced dipole-interaction

the stronger the bond , the higher the boiling point

therefore, N2, CO, H20

-------------------------------------->

Decreasing boiling point

Calculate [OH-] given [H3O+] in each aqueous solution and classify the solution as acidic or basic. [H3O+] = 2.6 x 10-8 M

Answers

Answer:

To calculate the [OH-] in the solution we must first find the pOH

That's

pH + pOH = 14

pOH = 14 - pH

First to find the pH we use the formula

pH = - log [H3O+]

From the question

[H3O+]= 2.6 × 10^-8 M

pH = - log 2.6 × 10^-8

pH = 7.6

pH = 8

So we pOH is

pOH = 14 - 8 = 6

To find the [OH-] we use the formula

pOH = - log [OH-]

6 = - log [OH-]

Find antilog of both sides

[OH-] = 1.0 × 10^-6 M

The solution is slightly basic since it's pH is in the basic region and slightly above the neutral point 7

Hope this helps you

11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. If a chicken weighs 3kg, how many grams of cyanide would it need to ingest to kill 50% of domestic chickens?

Answers

Answer:

[tex]0.033g[/tex]

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

[tex]11mg\longrightarrow 1kg\\\\?\ \ \ \ \ \ \longrightarrow 3kg[/tex]

Thus, we obtain:

[tex]?=\frac{3kg*11mg}{1kg}\\ \\?=33mg[/tex]

That in grams is:

[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]

Regards.

The total kinetic energy of a body is known as:

A. Thermal energy
B. Convection
C. Potential energy
D. Temperature

Answers

The total kinetic energy of a body is known as Thermal energy. Option A

What is thermal energy?

Thermal energy is the direct sum of all the available random kinetic energies of molecules.

Also note that thermal energy is directly proportional to temperature in Kelvin.

Thus, the total kinetic energy of a body is known as Thermal energy. Option A

Learn more about thermal energy here:

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Answer:

A.) Thermal energy

Explanation:

I got it correct on founders edtell

243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.​

Answers

Answer:

A. The atomic mass is 243 amu, and the atomic number is 95.

Solution of the Schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the behavior of the electron. Each function is characterized by 3 quantum numbers: n, l, and ml. If the value of n = 3 ... The quantum number l can have values from ? to ? . ... The total number of orbitals possible at the n = 3 energy level is ? . If the value of l = 3 ... The quantum number ml can have values from to ? . ... The total number of orbitals possible at the l = 3 sublevel is ?? .

Answers

Answer:

1) The quantum number l can have values from

2 to 0

2)The total number of orbitals possible at the n = 3 energy level is 3'2=9

3) If the value of l = 3 ... The quantum number ml can have values from 3 to -3

The quantum number l determines the shape of the orbital.  In this case, if the value of n is 3, then the quantum number l can have values from 0 to (3-1), which is 2.

The total number of orbitals possible at the n = 3 energy level can be determined using the formula 2l + 1. So, for l = 0, there is 1 orbital. For l = 1, there are 3 orbitals. And for l = 2, there are 5 orbitals. Therefore, the total number of orbitals possible at the n = 3 energy level is 1 + 3 + 5 = 9.

On the other hand, the quantum number ml represents the magnetic quantum number. It specifies the orientation of the orbital in space. The value of ml ranges from -l to +l. So, if the value of l is 3, then the quantum number ml can have values from -3 to +3.

The total number of orbitals possible at the l = 3 sublevel can be determined using the formula 2ml + 1. So, for ml = -3, there is 1 orbital. For ml = -2, there is 3 orbitals. For ml = -1, there is 5 orbitals. For ml = 0, there is 7 orbitals. For ml = 1, there is 5 orbitals. For ml = 2, there is 3 orbitals. And for ml = 3, there is 1 orbital.

Therefore, the total number of orbitals possible at the l = 3 sublevel is 1 + 3 + 5 + 7 + 5 + 3 + 1 = 25.

Learn more about quantum number,here:

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Consider the following practical aspects of titration.
(a) how can you tell when nearing the end point in titration?
(b) What volume of NaOH is required to permanently change the indicator at the end point?
(c) If KHP sample #1 requires 19.90 mL of NaOH solution to reach an end point, what volume is required for samples #2 and #3?
(d) if vinegar sample #1 requires 29.05 mL of NaOH solution to reach an endpoint, what volume is required for samole #2 and #3?

Answers

Answer:

A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration  

B) The volume of NaOH required to permanently change the indicator at the end point is  a  drop of NaOH

c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same

D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same

Explanation:

A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration  

B) The volume of NaOH required to permanently change the indicator at the end point is  a  drop of NaOH

c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same

D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same

A soft drink contains 63 g of sugar in 378 g of H2O. What is the concentration of sugar in the soft drink in mass percent

Answers

Answer:

[tex]\% m/m= 14.3\%[/tex]

Explanation:

Hello,

In this case, the by mass percent is computed as shown below:

[tex]\% m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%[/tex]

Whereas the solute is the sugar and the solvent the water, therefore, the concentration results:

[tex]\% m/m=\frac{63g}{63g+378g} *100\%\\\\\% m/m= 14.3\%[/tex]

Best regards.

Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite under these conditions is ·0.710J·g−1K−1 . Be sure your answer has the correct number of significant digits.

Answers

Answer:

7.23 J

Explanation:

Step 1: Given data

Mass of graphite (m): 566.0 mgInitial temperature: 5.2 °CFinal temperature: 23.2 °CSpecific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)

Q = 7.23 J

4NH3(g) 5O2(g)4NO(g) 6H2O(g) Using standard thermodynamic data at 298K, calculate the free energy change when 1.81 moles of NH3(g) react at standard conditions.

Answers

Answer:

-434.14 kJ

Explanation:

Step 1: Write the balanced equation

4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)

Step 2: Calculate the standard free energy change (ΔG°r) for the reaction

We will use the following expression.

ΔG°r = 4 mol × ΔG°f(NO(g)) + 6 mol × ΔG°f(H₂O(g)) - 4 mol × ΔG°f(NH₃(g)) - 5 mol × ΔG°f(O₂(g))

ΔG°r = 4 mol × (86.55 kJ/mol) + 6 mol × (-228.57 kJ/mol) - 4 mol × (-16.45 kJ/mol) - 5 mol × (0 kJ/mol)

ΔG°r = -959.42 kJ

Step 3: Calculate the standard free energy change for 1.81 moles of NH₃

959.42 kJ are released per 4 moles of NH₃.

[tex]\frac{-959.42 kJ}{4mol} \times 1.81mol = -434.14 kJ[/tex]

Which of the examples is potassium?
es )
A)
B)
B
C)​

Answers

Answer:

examples of things which contain potassium are:

green vegetables

root vegetables

fruits

potassium chloride

potassium sulphate

Explanation:

if you need a specific answer please send the options

Answer:

C

Explanation:

The answer is the one with 20 protons, 20 neutrons, and 6-8-8-2 electrons.

In the pictured cell, the side containing zinc is the Choose... and the side containing copper is the Choose... . The purpose of the N a 2 S O 4 NaX2SOX4 is to

Answers

Answer:

Zinc- anode

Copper- cathode

Sodium sulphate- salt bridge

Explanation:

A galvanic cell is an electrochemical cell in which electrical energy is produced by a spontaneous chemical reaction.

In the pictured galvanic cell, zinc is the anode since it looses electrons according to the reaction; Zn(s) -----> Zn^2+(aq) + 2e

Copper is the cathode as shown here; Cu^2+(aq) + 2e ----> Cu(s)

Sodium sulphate functions as the salt bridge. It keeps the both solutions neutral by ensuring charge balance in the both half cells.

Answer:

zinc=anode

copper=cathode

Explanation:

When 0.100 M NaOH is titrated with 25.00 mL 0.0500 M HBr, which of the following is correct for this titration?

A. Initially the pH will be less than 1.00.
B. The pH at the equivalence point will be 7.00.
C. It will require 12.50 mL of NaOH to reach the equivalence point.

When 0.100 M NaOH is titrated with 25.00 mL 0.0500 M HBr, which of the following is correct for this titration?

A. Initially the pH will be less than 1.00.
B. The pH at the equivalence point will be 7.00.
C. It will require 12.50 mL of NaOH to reach the equivalence point.

a)A, C
b) A, B
c) B, C
d) B

Answers

Answer:

c) B, C

Explanation:

NaOH(aq) + HBr(aq) -----> NaBr(aq) +H2O(l)

1) concentration of acid CA= 0.05 M

Concentration of base CB= 0.1 M

Volume of acid VA= 25.00ml

Volume of base VB= unknown

Number of moles of acid NA= 1

Number of moles of base NB= 1

CAVA/CBVB = NA/NB

CAVANB =CBVBNA

VB= CAVANB/CB NB

VB= 0.05 × 25 × 1/ 0.1 ×1

VB= 12.5 ML

2.

Which of the following represents a compound made of five molecules? CO 5 C 2O 5 C 5O 5CO 2

Answers

Answer:

Co5

Please mark me brainliest so that I get encouraged to make more great answers like this one!

Answer:

GUYS ITS 5CO 2

Explanation:

The literature value for the Ksp of Ca(OH)2 at 25 °C is 4.68E−6. Imagine you ran the experiment and got a calculated value for Ksp which was too high. Select all of the possible circumstances which would cause this result.

A. The HCl was more concentrated than the labeled molarity (0.0500 M).

B. The Ca[OH]2 solution may have been supersaturated.

C. The HCl was less concentrated than the labeled molarity (0.0500 M).

D. The Ca[OH]2 solution may have been unsaturated.

E. The titration flask may have not been clean and had a residue of a basic solution.

F. The titration flask may have not been clean and had a residue of an acidic solution.

Answers

Answer:

D. The Ca[OH]2 solution may have been unsaturated

Explanation:

The solubility product constant Ksp of any given chemical compound is a term used to describe the equilibrium between a solid and the ions it contains solution. The value of the Ksp indicates the extent to which any compound can dissociate into ions in water. A higher the Ksp, implies more greater solubility of the compound in water.

If the Ksp is more than the value in literature, this false value must have arisen from the fact that the solution was unsaturated hence it appears to be more soluble than it should normally be when saturated.

Using only sodium carbonate, Na2CO3, sodium bicarbonate, NaHCO3, and distilled water determine how you could prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3. The total molarity of the ions should be 0.20 M. The Ka of the hydrogen carbonate ion, HCO3 - , is 4.7 x 10-11 .

Answers

Answer:

Weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.

Explanation:

The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation:

[tex] pH = pKa + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]  

We have that pH = 10.3 and the Ka is 4.7x10⁻¹¹, so:

[tex] 10.3 = -log(4.7 \cdot 10^{-11}) + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]  

[tex] \frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]  (1)

Also, we know that:

[tex] [Na_{2}CO_{3}] + [NaHCO_{3}] = 0.20 M [/tex]    (2)

From equation (2) we have:

[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] [/tex]   (3)

By entering (3) into (1):

[tex] \frac{0.20 - [NaHCO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]

[tex] 0.94*[NaHCO_{3}] + [NaHCO_{3}] = 0.20 [/tex]

[tex] [NaHCO_{3}] = 0.103 M [/tex]  

Hence, the [Na_{2}CO_{3}] is:

[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] = 0.20 M - 0.103 M = 0.097 M [/tex]  

Now, having the concentrations and knowing the volume of the buffer solution we can find the mass of the sodium carbonate and the sodium bicarbonate, as follows:

[tex]m_{Na_{2}CO_{3}} = C*V*M = 0.097 mol/L*0.050 L*105.99 g/mol = 0.5141 g[/tex]

[tex]m_{NaHCO_{3}} = C*V*M = 0.103 mol/L*0.050 L*84.007 g/mol = 0.4326 g[/tex]

Therefore, to prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3 we need to weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.      

   

I hope it helps you!

discuss four factors of learning ​

Answers

Answer:

plz mark as BRAINLIEST plz...

Explanation:

●Intellectual factor: The term refers to the individual mental level. ...

●Learning factors: ...

●Physical factors: ...

●Mental factors: ...

When balancing redox reactions under acidic conditions, hydrogen is balanced by adding: Select the correct answer below:
a. hydrogen gas
b. water molecules
c. hydrogen atoms
d. hydrogen ions

Answers

Answer:

water molecules

Explanation:

Redox reactions are carried out under acidic or basic conditions as the case may be.

If the reaction is carried out in an acid medium, then we must balance the hydrogen ions on the lefthand side of the reaction equation with water molecules on the righthand side of the reaction equation.

For instance, the equation for reduction of MnO4^- under acidic condition is shown below;

MnO4^-(aq) + 5e + 8H^+(aq) --------> Mn^2+(aq) + 4H2O(l)

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