What is 2KNO3 + 1H2CO3 = 1K2CO3+2HNO3 reaction type

WARRIOR [948]

the answer is d electronic balance

8 0

3 years ago

Hydrogen sulfide (H2S) is a gas that smells like rotten eggs. It can be produced by bacteria in your mouth and contributes to ba

Doss [256]

Answer:

See explanation.

Explanation:

Hello!

In this case, we consider the questions:

a. Ideal gas at:

i. 273.15 K and 22.414 L.

ii. 500 K and 100 cm³.

b. Van der Waals gas at:

i. 273.15 K and 22.414 L.

ii. 500 K and 100 cm³.

Thus, we define the ideal gas equation and the van der Waals one as shown below:

$P^{id}=\frac{nRT}{V}\\\\ P^{vdW}=\frac{RT}{v-b}-\frac{a}{v^2}$

Whereas b and a for hydrogen sulfide are 0.0434 L/mol and 4.484 L²*atm / mol² respectively, therefore, we proceed as follows:

a.

i. 273.15 K and 22.414 L.

$P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L}=1 .00 atm$

ii. 500 K and 100 cm³ (0.1 L).

$P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*500K}{0.100L}=410.3 atm$

b.

i. 273.15 K and 22.414 L: in this case, v = 22.414 L / 1.00 mol = 22.414 L/mol

$P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(22.414L/mol)^2}=0.993atm$

ii. 500 K and 100 cm³: in this case, v = 0.1 L / 1.00 mol = 0.100 L/mol

$P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*500K}{0.100L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(0.100L/mol)^2}=276.5atm$

Whereas we can see a significant difference when the gas is at 500 K and occupy a volume of 0.100 L.

Best regards!

5 0

3 years ago

9. Calculate the percentage composition of silver sulfate, Ag2SO4.

riadik2000 [5.3K]

Answer:

69% Ag, 10.5% S, 20.5% O.

Explanation:

Ag₂SO₄ → Molar mass = 311.8 g/mol

2 mol of Ag → 107.87g/m . 2m = 215.74 g

1 mol of S → 32.06g/m . 1m = 32.06 g

4 mol of O → 16g/m .4m = 64 g

Percentage composition will be:

(Mass of Ag/ Total mass) . 100

(215.74 g / 311.8 g) . 100 = 69 %

(Mass of S/ Total mass) . 100

(32.06 g / 311.8 g) .100 = 10.5%

(Mass of O/ Total mass) . 100

(64 g / 311.8 g) . 100 = 20.5 %

7 0

3 years ago

In which reaction does the oxidation number of hydrogen change? In which reaction does the oxidation number of hydrogen change?

dedylja [7]

Answer: The correct answer is $2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)$

Explanation:

Oxidation number is defined as the number which is given to an atom when it looses or gains electron. When an atom looses electron, it attains a positive oxidation state. When an atom gains electron, it attains a negative oxidation state.

Oxidation state of the atoms in their elemental state is considered as 0. Hydrogen is present as gaseous state.

For the given chemical reactions:

Reaction 1: $2HClO_4(aq.)+CaCO_3(s)\rightarrow Ca(ClO_4)_2(aq.)+H_2O(l)+CO_2 (g)$