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2 years ago

I attached my homework sheet, and can't figure it out. If anyone could help it would be great.

Chemistry

1 answer:

Lynna [10]2 years ago

7 0

Answer:

I can't see the picture can you just text it to me so I can help you

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PLEASEEE HELP ITS DUE SOON!

erastovalidia [21]

Answer:

The correct answer would be C) Support

4 0

3 years ago

Which of the following acids is in household vinegar? A nitric acid hydrochloric acid C fluoric acid D acetic acid

Free_Kalibri [48]

Answer: Out of the given options acetic acid is in household vinegar.

Explanation:

Household vinegar is the one that is commonly used in our home while cooking a number of dishes.

The common name of vinegar is acetic acid and its chemical formula is $CH_{3}COOH$ .

For example, vinegar (acetic acid) is used while making noodles.

Thus, we can conclude that out of the given options acetic acid is in household vinegar.

7 0

3 years ago

CH4 + 202 → CO2 + 2H2O

ohaa [14]

It’s a combustion reaction but what are you trying to find?

6 0

2 years ago

Acetic acid is the active ingredient in vinegar. In a solution of acetic acid the following equilibrium is established. HC2H3O2(

lana [24]

Answer:

the equilibrium constant is 1.8 x 10⁻5 and strongly favor the reactants.

Explanation:

the chemical reaction provided for the two equation are the same but different direction i.e a reversible reaction. Assuming, the mass of reactants and product and temperature remain constant.

therefore, the equilibrium constant K, is 1.8 x 10⁻5. this is a very small value of K, thereby strongly favor the backward direction to form reactant.

6 0

3 years ago

A mixture of hydrochloric and sulfuric acids is prepared so that it contains 0.315 M HCl and 0.125 M H2SO4. What volume of 0.55

Xelga [282]

Answer: The volume of NaOH required is 402.9 mL

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For HCl:

Molarity of HCl solution = 0.315 M

Volume of solution = 503.4 mL = 0.5034 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.315M=\frac{\text{Moles of HCl}}{0.5034L}\\\\\text{Moles of HCl}=(0.315mol/L\times 0.5034L)=0.1586mol$

For sulfuric acid:

Molarity of sulfuric acid solution = 0.125 M

Volume of solution = 503.4 mL = 0.5034 L

Putting values in equation 1, we get:

$0.125M=\frac{\text{Moles of }H_2SO_4}{0.5034L}\\\\\text{Moles of }H_2SO_4=(0.125mol/L\times 0.5034L)=0.0630mol$

As, all of the acid is neutralized, so moles of NaOH = [0.1586 + 0.0630] moles = 0.2216 moles

Molarity of NaOH solution = 0.55 M

Moles of NaOH = 0.2216 moles

Putting values in equation 1, we get:

$0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL$

Hence, the volume of NaOH required is 402.9 mL

6 0

3 years ago