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Shalnov [3]
2 years ago
9

You have 350 mL of 3.4 M hydrochloric acid (HCl). How many grams of HCl gas are dissolved? Bonus: what is the volume of the HCl

at STP?
Chemistry
1 answer:
Crank2 years ago
5 0

Answer:

1. 43.44g of HCl

2. 26.67 L of HCl

Explanation:

1) Molarity of a solution = number of moles (n) ÷ Volume (V)

According to the provided information in this question,

V = 350 mL = 350/1000 = 0.350L

Molarity = 3.4 M

Using Molarity = n/V

3.4 = n/0.350

n = 3.4 × 0.350

n = 1.19mol

Using the formula below to calculate the mass of HCl;

mole = mass/molar mass

Molar mass of HCl = 1 + 35.5 = 36.5g/mol

mole = mass/MM

mass = 1.19 mol × 36.5g/mol

mass = 43.44g of HCl

2) At STP, HCl has a pressure of 1atm, a temperature of 273K

V = ?

n = 1.19 mol

R = 0.0821 Latm/molK

Using PV = nRT

V = nRT/P

V = 1.19 × 0.0821 × 273/1

Volume = 26.67L

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For example, lets say t = pink nose and T = black nose.

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But, they also have hidden pink nose genotypes.

These hidden genotypes are passed on to the puppies, one of the puppies gets more black nose genotypes.

If a puppy gets tt, they will have a pink nose.

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3 years ago
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Easy way to remember the difference between ionic and covalent bonds?
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Answer:

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1 year ago
The equilibrium constant for the reaction 2NO2(g) N2O4(g) is Keq = . If a sample at equilibrium was found to contain 0.058 M NO2
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Answer:

\boxed{3.6}

Explanation:

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E/mol·L⁻¹:   0.058     0.012

K_{\text{eq}} = \dfrac{\text{[N$_{2}$O$_{4}$]}}{\text{[NO$_{2}$]$^{2}$}} = \dfrac{0.012}{0.058^{2}} = \mathbf{3.6} \\\\

\text{The $K_{\text{eq}}$ value would be $\boxed{\mathbf{3.6}}$}

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