Ask question

Ask question

All categories

4 years ago

13

Determine the molarity for the fixer Sodium carbonate decahydrate (I dissolved 127.62 grams of sodium carbonate for every 4 lite

rs of water)

1 answer:

grin007 [14]4 years ago

4 0

Answer:

0.112 M.

Explanation:

Molarity is the no. of moles of solute in a 1.0 L of the solution.

M = n/V.

<em>M = (mass/molar mass)solute x (1000/V of the solution).</em>

mass = 127.62 g.

molar mass = 286.138 g/mol.

V of the solution = 4.0 L = 4000.0 mL.

<em>∴ M = (mass/molar mass)solute x (1000/V of the solution)</em> = (127.62 g / 286.138 g/mol) x (1000 / 4000.0 mL) = <em>0.1115 M ≅ 0.112 M.</em>

You might be interested in

Moving water carrying away small

kondaur [170]

Answer:

erosion:)

Explanation:

7 0

3 years ago

Read 2 more answers

Choose all the answers that apply.

Lady_Fox [76]

It is softer than topaz and and it is softer than diamond (diamond has a Mohs hardness of 10, which is the highest value of the scale)

7 0

3 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

Identify the conjugate acid/base pairs in each of the following equations:

Valentin [98]

Answer:

(a) Pair 1: H₂S and HS⁻

Pair 2: NH₃ and NH₄⁺

(b) Pair 1: HSO₄⁻ and SO₄⁻

Pair 2: NH₃ and NH₄⁺

(c) Pair 1: HBr and Br⁻

Pair 2: CH₃O⁻ and CH₃OH

(d) Pair 1: HNO₃ and NO₃⁻

Pair 2: H₃O⁺

Explanation:

When an acid loses its proton (H⁺), a conjugate base is produced.

When a base accepts a proton (H⁺), it forms a conjugate acid.

(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.

NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺

(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.

The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.

(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.

CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.

(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.

H₂O gains a proton to form the conjugate acid H₃O⁺.

6 0

3 years ago

In order to change a solid to a liquid or a liquid to a gas, what is needed?

Neporo4naja [7]

Heat; rather, or change of the molecules to make them move faster

6 0

3 years ago