Which statement is true about dissolving?

Calclulate the mass of one molecule of nitrogen in gram

Mama L [17]

Answer:

for its one mole it is of 28gram. according to mole concept in one mole, there is 6.022*10^23 moles are present. so mass of 1 molecule of nitogen gas is 28/(6.022*10^23)

8 0

3 years ago

Read 2 more answers

Type Calculations. Given the balanced equation: 2 Al + 3 H2SO4---> Al2(SO4)3 + 3 H2 Molar mass (g/mol): Al=26.98; H2SO4=98.08

Oxana [17]

Answer:

A. 1.88 mol H₂

B. 182 g Al₂(SO₄)₃

C. 54.8%

Explanation:

2 Al + 3 H₂SO₄ ⇒ Al₂(SO₄)₃ + 3 H₂

A. Convert grams of Al to moles. The molar mass is 26.98 g/mol.

(33.8 g)/(26.98 g/mol) = 1.253 mol Al

Use stoichiometry to convert moles of Al to moles of H₂. Looking at the equation, you can see that for every 2 mol of Al consumed, 3 moles of H₂ is produced. Use this relationship.

(1.253 mol Al) × (3 mol H₂)/(2 mol Al) = 1.879 mol H₂

You will produce 1.88 mol of H₂ gas.

B. Again, use stoichiometry. For every 3 moles of H₂SO₄ consumed, 1 mole of Al₂(SO₄)₃ is produced.

(1.60 mol H₂SO₄) × (1 mol Al₂(SO₄)₃/3 mol H₂SO₄) = 0.533 mol Al₂(SO₄)₃

Convert moles of Al₂(SO₄)₃ to grams. The molar mass is 342.15 g/mol.

(0.533 mol) × (342.15 g/mol) = 182.48 g Al₂(SO₄)₃

You will produce 182 g of Al₂(SO₄)₃.

C. Calculate percent yield by dividing the actual yield by the theoretical yield. Multiply by 100%.

(100.0/182.48) × 100% = 54.8%

The percent yield is 54.8%.

7 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

4 years ago

A vial containing radioactive selenium-75 has an activity of 2.3 mci/ml. If 2.6 mci are required for a leukemia test, how many m

nika2105 [10]

Answer is: 1130 microliters must be administered.

activity(selenium-75) = 2.3 mCi/mL.

radioactivity(Se-75) = 2.6 mCi; mCi is millicurie.

The curie (symbol: Ci) is a non-SI unit of radioactivity.

V(Se-75) = radioactivity(Se-75) ÷ activity(Se-75).

V(Se-75) = 2.6 mCi ÷ 2.3 mCi/mL.

V(Se-75) = 1.13 mL; volume.

V(Se-75) = 1.13 mL · 1000µL/mL.

V(Se-75) = 1130 µL.

3 0

3 years ago

29) What type of chemical reaction is given by the following balanced equation: Al2O3 → Al + O2

Law Incorporation [45]

Decomposition.

Decomposition is the chemical reaction in which a compound breaks down into the two elements or molecules that formed the compound. In this case, Dialuminium Trioxide has broken down into the basic components of Al2 and O2.

Hope this helps!

7 0

3 years ago