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alexandr1967 [171]
3 years ago
15

Which of the following is an example of a chemical change?

Chemistry
1 answer:
elena-s [515]3 years ago
6 0
The best answer is the first option. A bike rusting is an example of a chemical change since rusting is a chemical reaction. New substances are formed from this reaction. On the other hand, other choices given are just physical changes because only the shape or the physical form of the substances are being changed.
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How many particles are in a 151 g sample of Li2O?
neonofarm [45]

Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

To convert this mass value to moles, we use;

moles = mass/molar mass

Molar mass of Li2O = 6.9(2) + 16

= 13.8 + 16

= 29.8g/mol

Mole = 151/29.8g

mole = 5.07moles

number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23

= 30.52 × 10^23

= 3.052 × 10^24 particles.

4 0
3 years ago
Carbon dioxide is released into the atmosphere by _____.
kirill115 [55]

Answer:

4 a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation.;

Explanation:

a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation

3 0
2 years ago
Carbon monoxide (CO) reacts with hydrogen (H2) to form methane (CH4) and water (H2O).
Artemon [7]

Answer:

5.9x10^-2 M

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Concentration of CO, [CO] = 0.30 M

Concentration of H2, [H2] = 0.10 M

Concentration of H2O, [H2O] = 0.020 M

Equilibrium constant, K = 3.90

Concentration of CH4, [CH4] =..?

Step 2:

The balanced equation for the reaction. This is given below:

CO(g) + 3H2(g) <=> CH4(g) + H2O(g)

Step 3:

Determination of the concentration of CH4.

The expression for equilibrium constant of the above equation is given below:

K = [CH4] [H2O] / [CO] [H2]^3

3.9 = [CH4] x 0.02/ 0.3 x (0.1)^3

Cross multiply to express in linear form

[CH4] x 0.02= 3.9 x 0.3 x (0.1)^3

Divide both side by 0.02

[CH4] = 3.9 x 0.3 x (0.1)^3 /0.02

[CH4] = 5.9x10^-2 M

Therefore, the equilibrium concentration of CH4 is 5.9x10^-2 M

5 0
3 years ago
A sample of a chromium-containing alloy weighing 3.450 g was dissolved in acid, and all the chromium in the sample was oxidized
Zina [86]

Answer:

H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-

Explanation:

Reduction half reaction

2H_2O + CrO_4^2- + 3e -> CrO_2^- + 4OH^-

Oxidation half reaction

2OH^- + SO_3^2- -> SO_4^2- + H_2O + 2e

Balanced overall equation

H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-

 

5 0
3 years ago
Oxygen in the air can react with wood when a match is burned. Ash is produced.
Hitman42 [59]

Answer:

D. Ash is a different substance than wood.

8 0
2 years ago
Read 2 more answers
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