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2 years ago

Which of the following is an example of a decomposition reaction?

Chemistry

2 answers:

Ivenika [448]2 years ago

7 0

Answer:

C is your answer.

Explanation:

A decomposition reaction occurs when a single product breaks down into two or more simpler products.

AB → A + B

Veronika [31]2 years ago

6 0

Answer:

C. 2NaCl → 2Na+ Cl2

Explanation:

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How many molecules of carbon dioxide are dissolved in 0.550 L of water at 25 °C if the pressure of CO2 above the water is 0.250

Grace [21]

Answer: The number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.034mol/L.atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = pressure of carbon dioxide gas = 0.250 atm

Putting values in above equation, we get:

$C_{CO_2}=0.034mol/L.atm\times 0.250atm\\\\C_{CO_2}=8.5\times 10^{-3}M$

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

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

Molarity of carbon dioxide = $8.5\times 10^{-5}M$

Volume of solution = 0.550 L

Putting values in above equation, we get:

$8.5\times 10^{-3}M=\frac{\text{Moles of }CO_2}{0.550L}\\\\\text{Moles of }CO_2=(8.5\times 10^{-3}mol/L\times 0.550L)=4.675\times 10^{-3}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules

So, $4.675\times 10^{-3}$ moles of carbon dioxide will contain = $(6.022\times 10^{23}\times 4.675\times 10^{-3})=2.815\times 10^{21}$ number of molecules

Hence, the number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

3 0

3 years ago

Read 2 more answers

The half life for the decay of carbon- is years. Suppose the activity due to the radioactive decay of the carbon- in a tiny samp

EleoNora [17]

The question is incomplete. The complete question is:

The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.

Answer:

570 years

Explanation:

The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.

6 0

3 years ago

How does geothermal energy differ from solar energy?

Crazy boy [7]

Answer:

Whereas solar energy makes use of the sun to generate energy, geothermal energy makes use of the heat that is trapped deep in the center of the earth. The magma buried deep down of our soil is as hot as the sun’s surface, and some of that heat manages to escape outward. When that happens, we can harness that said heat for energy.

Explanation:

8 0

3 years ago

Read 2 more answers

What is insulin's structure? a.quaternary b.secondary c.primary d.tertiary

svetoff [14.1K]

Answer:

Explanation:

Quaternary structure of proteins is composed of two or more polypeptide chains. Insulin has two; one alpha and one beta chain. The two chains are joined together by disulfide bonds at two points (at cysteines). Other examples of quaternary proteins structures are DNA polymerase and hemoglobin.

7 0

2 years ago

Which of the following statements is true about the specific heat capacity of a substance?

Ludmilka [50]

Answer:

Option A

Explanation:

An intensive property is a bulk property, meaning that it is a local physical property of a system that does not depend on the system size or the amount of material in the system. Examples of intensive properties include temperature, T; refractive index, n; density, ρ; and hardness of an object,specific heat, η.

Physical properties can be observed or measured without changing the composition of matter. Physical properties are used to observe and describe matter. Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, specific heat and many others.

6 0

3 years ago