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dalvyx [7]
2 years ago
11

Which chemical equation is a model of a decomposition reaction?

Chemistry
1 answer:
blagie [28]2 years ago
3 0
A is a model of a decomposition reaction
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All of these are correct.
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2 years ago
The diagram below is an artist’s impression of a single atom of element Be. The neutrons are shown with stripes, the protons are
Ahat [919]

Answer is: beryllium-10.

The diagram shows that atom has 4 protons, 5 neutrons and 2 valence electrons.

Atomic number is the number of protons, which is characteristic of a chemical element, beryllium (Be) is an element with atomic number 4.

Two valence electrons means that atom is from 2. group of periodic table, only beryllium is from that group; sodium (1. group), boron (13. group) and carbon (14. group).

Beryllium-10 has 6 neutrons, so it is isotope (different number of neutrons or mass number).

3 0
3 years ago
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How many liters of salt solution will be needed to provide 15 grams of salt if the concentration of the solution is 50 g/l?​
asambeis [7]

Answer:

.30 l

Explanation:

15g/50g/l= .3l

6 0
3 years ago
If a sodium ion has 11 protons, 12 neutrons, and 11 electrons, what is the atomic mass of the atom?
iragen [17]

Answer:

23

Explanation:

we do not care about electrons, so 11 + 12 = 23

7 0
3 years ago
1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
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