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3 years ago

A chemist obtains a 0.300 M solution of HCI and adds 10.0 mL of the HCI to CaCO3 how many moles of HCI did she add?

Chemistry

1 answer:

Salsk061 [2.6K]3 years ago

4 0

Answer:

Number of moles = 0.003 mol

Explanation:

Given data:

Molarity of HCl = 0.300 M

Volume = 10.0 mL (0.01 L)

Number of moles of HCl added = ?

Solution:

Formula:

Molarity = number of moles / volume in litter

Number of moles = Molarity × volume in litter

Number of moles = 0.300 mol/L × 0.01 L

Number of moles = 0.003 mol

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3 years ago

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12 Significant Figures in Addition and

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3 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

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3 years ago

In the force equation, which two variables have a direct relationship? Which two have an indirect/inverse relationship?

zhuklara [117]

Answer:

if the dependent and independent variable go up its direct but if one of them goes up and another goes down its indirect

Explanation:

for example the more mass an object has the less speed it has that's indirect but the less salt you add to water the less time you have to wait for it to boil

i hope this was useful

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3 years ago