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GarryVolchara [31]
4 years ago
15

What is not true of a strong acid?

Chemistry
2 answers:
BartSMP [9]4 years ago
7 0

Answer:The correct answer is option B.

Explanation:

Acids are those chemical compounds which gives free H^+ ions in their aqueous solution.

Strong acids are:

  • They strong electrolyte that is they completely dissociates in water.
  • The strongest acids have concentration lower than 1.0 M.
  • They conduct electricity in the solution.

From the above enlisted properties of acid the correct answer is option B.

34kurt4 years ago
3 0
B. It always has a concentration above 1.0 M
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<span>The balance format is
4NH3+ 5O2 -------> 4NO + 6H2O </span>
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Grady is doing an experiment about the solubility of sugar. He puts 100 milliliters of water in each of three beakers. He leaves
olga nikolaevna [1]

Grady is doing an experiment about the solubility of sugar. He puts 100 milliliters of water in each of three beakers. He leaves the first beaker at room temperature, heats the second beaker to 60°C, and heats the third beaker until the water boils at 100°C. The variable Grady change on  purpose in the experiment is the temperature of water in each beaker .

Variables in the experiment is the any factor that can exist in different types or amount. There are three types of variables: independent variable , dependent variable , controlled variable. The independent variable is the variable you changed in the experiment. dependent variable is that changes because of independent variable. the controlled variable is the constant one.

Thus, Grady is doing an experiment about the solubility of sugar. He puts 100 milliliters of water in each of three beakers. He leaves the first beaker at room temperature, heats the second beaker to 60°C, and heats the third beaker until the water boils at 100°C. The variable Grady change on  purpose in the experiment is the temperature of water in each beaker .

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8 0
1 year ago
A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben
Anit [1.1K]

<u>Answer:</u> The percent yield of the reaction is 32.34 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For benzoic acid:</u>

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol

The chemical equation for the reaction of benzoic acid and methanol is:

\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = \frac{1}{1}\times 0.0295=0.108 moles of methyl benzoate

  • Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g

  • To calculate the percentage yield of methyl benzoate, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%

Hence, the percent yield of the reaction is 32.34 %

6 0
3 years ago
Evidence that best supports the theory of biological evolution was obtained from the
Alisiya [41]
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3 years ago
A solution of acetic acid has a pH of 3.45. What is the concentration of acetic acid in this solution? Ka for CH3COOH is 1.8×10-
Serjik [45]

Answer:

0.007 M

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Thus,  

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The expression of the pH of the calculation of weak acid is:-

pH=-log(\sqrt{k_a\times C})

Where, C is the concentration = ?

Given, pH = 3.45

So, for CH_3COOH, K_a=1.8\times 10^{-5}

3.45=-log(\sqrt{1.8\times 10^{-5}\times C})

\log _{10}\left(\sqrt{1.8\cdot \:10^{-5}C}\right)=-3.45

C = 0.007 M

5 0
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