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Vera_Pavlovna [14]
3 years ago
8

What are the carbon (C) and oxygen (O) in carbon dioxide (CO2)?

Chemistry
1 answer:
Fynjy0 [20]3 years ago
6 0
A. Atoms. Because I learned this in 6th grade.
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Which of the following describes the energy transformations that occur when a toaster is used to make toast?
denis23 [38]

Answer:

Thermal energy (B)

Explanation:

just did it

8 0
3 years ago
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A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction
Molodets [167]

<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

<u>Explanation:</u>

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)

<u>For NaOH:</u>

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L   (Conversion factor:   1 L = 1000 mL)

Putting values in equation 1, we get:

0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol

<u>For sulfuric acid:</u>

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol

The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 2.925\times 10^{-3} moles of KOH will react with = \frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = (4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, 2.925\times 10^{-3} moles of KOH will produce = \frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol of sodium sulfate

  • <u>For sodium sulfate:</u>

Moles of sodium sulfate = 1.462\times 10^{-3}moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M

  • <u>For sulfuric acid:</u>

Moles of excess sulfuric acid = 3.498\times 10^{-3}mol

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M

  • <u>For NaOH:</u>

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

\text{Molarity of NaOH}=\frac{0}{0.050}=0M

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0
4 years ago
Perform the following
irinina [24]

Answer:

12.0778

Explanation:

7 0
3 years ago
What is the molarity of a solution containing 8.0 moles of solute in 120mL of solution?
Nataliya [291]
Answer:
200/3 M which is approximately equal to 66.6667 M

Explanation:
Molarity is defined as the number of moles of solute per liter of solution.

It can be calculated as follows:
molarity =  \frac{number of moles of solute}{liters of solution}

We are given that:
number of moles of solute = 8 moles
volume of solution = 120 ml = 0.12 liters

Substitute with the givens in the above equation to get the molarity as follows:
molarity = \frac{8}{0.12} =  \frac{200}{3} = 66.6667 M

Hope this helps :)
3 0
3 years ago
The chemical formula of lead sulfide tells you that it contains<br>​
ra1l [238]

Answer:

sulfur oxygen

Explanation:

because its contain ide

7 0
4 years ago
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