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skad [1K]
3 years ago
15

What volume of 0.500 M HNO3(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?

Chemistry
1 answer:
asambeis [7]3 years ago
6 0
KOH+ HNO3--> KNO3+ H2O<span>
From this balanced equation, we know that 1 mol HNO3= 1 mol KOH (keep in mind this because it will be used later).

We also know that 0.100 M KOH aqueous solution (soln)= 0.100 mol KOH/ 1 L of KOH soln (this one is based on the definition of molarity).

First, we should find the mole of KOH:
100.0 mL KOH soln* (1 L KOH soln/ 1,000 mL KOH soln)* (0.100 mol KOH/ 1L KOH soln)= 1.00*10^(-2) mol KOH.

Now, let's find the volume of HNO3 soln:
1.00*10^(-2) mol KOH* (1 mol HNO3/ 1 mol KOH)* (1 L HNO3 soln/ 0.500 mol HNO3)* (1,000 mL HNO3 soln/ 1 L HNO3 soln)= 20.0 mL HNO3 soln.

The final answer is </span>(2) 20.0 mL.<span>

Also, this problem can also be done by using dimensional analysis. 

Hope this would help~ </span>
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Explanation

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5 0
3 years ago
State the definition of the partial molar Gibbs energy.
balu736 [363]

Explanation :

As we know that the Gibbs free energy is not only function of temperature and pressure but also amount of each substance in the system.

G=G(T,P,n_1,n_2)

where,

n_1\text{ and }n_2 is the amount of component 1 and 2 in the system.

Partial molar Gibbs free energy : The partial derivative of Gibbs free energy with respect to amount of component (i) of a mixture when other variable (T,P,n_j) are kept constant are known as partial molar Gibbs free energy of i^{th} component.

For a substance in a mixture, the chemical potential (\mu) is defined as the partial molar Gibbs free energy.

The expression will be:

\bar{G_i}=\mu_i=\frac{\partial G}{\partial n_i}_{(T,P,n_j)}

where,

T = temperature

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n_i\text{ and }n_j is the amount of component 'i' and 'j' in the system.

4 0
3 years ago
Find the number of moles of barium iodide if you have 5.23 x 1024 formula units of Bal2
Oksi-84 [34.3K]

Answer:

8.68 moles of BaI₂

Explanation:

Given data:

Number of moles of BaI₂ = ?

Number of formula units = 5.23× 10²⁴

Solution:

By using Avogadro number,

1 mole of any  substance contain 6.022× 10²³ formula units.

5.23× 10²⁴ formula units of BaI₂ × 1 mol / 6.022× 10²³ formula units

0.868 × 10¹ moles of BaI₂

8.68 moles of BaI₂

Thus, 5.23× 10²⁴ formula units of BaI₂ contain 8.68 moles of BaI₂

7 0
2 years ago
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