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3 years ago

What quantity in moles of HCl are there in 35.0 mL of 0.350 M HCl?

Chemistry

1 answer:

marta [7]3 years ago

5 0

Answer:

0.0123 moles

Explanation:

Concentration = Moles / Volume of solution

or you can rearrange the formula to get

Moles = concentration (moles/liter) x volume of solution (liter)

First convert your volume to L instead of mL. 35mL = 0.035L

moles = 0.350 moles/liter x 0.035 liter (liters cancel out)

moles = 0.0123

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Dolomite is a mixed carbonate of calcium and magnesium. Calcium and magnesium carbonates both decompose upon heating to produce

Setler79 [48]

Answer:

72.03 %

Explanation:

Total mass of dolomite = 9.66 g

Let the mass of Magnesium carbonate = x g

The mass of calcium carbonate = 9.66 - x g

Calculation of the moles of Magnesium carbonate as:-

Molar mass of Magnesium carbonate = 122.44 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{x\ g}{84.3139\ g/mol}=\frac{x}{84.3139}\ mol$

Calculation of the moles of calcium carbonate as:-

Molar mass of calcium carbonate = 100.0869 g/mol

Thus,

$Moles= \frac{9.66 - x\ g}{100.0869\ g/mol}=\frac{9.66 - x}{100.0869}\ mol$

According to the reaction shown below:-

$MgCO_3\rightarrow MgO+CO_2$

$CaCO_3\rightarrow CaO+CO_2$

In both the cases, the oxides formed from the carbonates in the 1:1 ratio.

So, Moles of MgO = $\frac{x}{84.3139}\ mol$

Molar mass of MgO = 40.3044 g/mol

Thus, Mass = Moles*Molar mass = $\frac{x}{84.3139}\times 40.3044 \ g$

Moles of CaO = $\frac{9.66 - x}{100.0869}\ mol$

Molar mass of CaO = 56.0774 g/mol

Thus, Mass = Moles*Molar mass = $\frac{9.66 - x}{100.0869}\times 56.0774 \ g$

Given that total mass of the oxide = 4.84 g

Thus,

$\frac{x}{84.3139}\times 40.3044 +\frac{9.66 - x}{100.0869}\times 56.0774=4.84$

$\frac{40.3044x}{84.3139}+56.0774\times \frac{-x+9.66}{100.0869}=4.84$

$-694.1618435x+45673.48749\dots =40843.38968\dots$

$x=\frac{4830.09780\dots }{694.1618435}$

$x=6.9582$

Thus, the mass of Magnesium carbonate = 6.9582 g

$\%\ mass=\frac{Mass_{MgCO_3}}{Total\ mass}\times 100$

$\%\ mass=\frac{6.9582}{9.66}\times 100=72.03\ \%$

3 0

3 years ago

Now consider the reaction a+2b⇌c for which in the initial mixture qc=[c][a][b]2=387 is the reaction at equilibrium? if not, in w

Paul [167]

Solving this chemistry is a little bit hard because the question didn't give some important detailed.
So first, there are a couple problems with your question.
We will just need to know which direction will it proceed to reach equilibrium.
Your expression for Kc (and Qc ) for the reaction should be:
Kc = [C] / [A] [B]^2
You have not provided a value for Kc, so a value of Qc tells you absolutely nothing. Qc is only valuable in relation to a numerical value for Kc. If Qc = Kc, then the reaction is at equilibrium. If Q < K, the reaction will form more products to reach equilibrium, and if Q > Kc, the reaction will form more reactants.

5 0

4 years ago

If the copper cube had a mass of 28.7 grams and a volume of 3.2 mL, what would its density be?

Otrada [13]

Answer:

ρ=m÷v

=28.7÷3.2

≈8.97 g/m³

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3 years ago

Choose all the answers that apply.

Arada [10]

Answer:

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Explanation:

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Answer:

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