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Anastasy [175]
3 years ago
12

When two gases of a chemical reaction are both at the same temperature and pressure and they have the same molar volume, what is

their stoichiometric ratio in the chemical equation describing the reaction?
Chemistry
2 answers:
Rudik [331]3 years ago
4 0
When two gases of a chemical reaction are at the same temperature, pressure and molar volume, then the stoichiometric ratio of the gases would be 1 is to 1. Molar volume is the volume of the gas per mole of the gas. Having the same conditions for both gases would mean that they are present with the same number of moles.
monitta3 years ago
3 0

Answer:

a. 1 to 1

Explanation:

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Margaret [11]
As mentioned above, phosphoric acid has 3 pKa values, and after 3 ionization it gives 3 types of ions at different pKa values:

H₃PO₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq)         pKₐ₁ 
<span>

</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq)       pKₐ₂


HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq)          pKₐ₃ 


At the highest pKa value (12.4) of phosphoric acid, the last OH group will lose its hydrogen. On the picture I attached, it is shown required protonated form of phosphoric acid before reaction whose pKa value is 12.4.


6 0
3 years ago
A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate
kotegsom [21]

<u>Answer:</u> The final temperature of water is 32.3°C

<u>Explanation:</u>

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of solution 1 (liquid water) = 50.0 g

m_2 = mass of solution 2 (liquid water) = 29.0 g

T_{final} = final temperature = ?

T_1 = initial temperature of solution 1 = 25°C  = [273 + 25] = 298 K

T_2 = initial temperature of solution 2 = 45°C  = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K

Converting this into degree Celsius, we use the conversion factor:

T(K)=T(^oC)+273

305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC

Hence, the final temperature of water is 32.3°C

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