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motikmotik
3 years ago
15

How many carbon atoms are in each mole of calcium carbonate?

Chemistry
1 answer:
xeze [42]3 years ago
6 0

Answer:

5 atoms are in each molecule of calcium carbonate

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Can someone please help me in this question?
Monica [59]

Answer: the beaker with a higher boiling point is B.

Explanation:

since it contains more mass/ volume it takes up more time to heat up.

hope this helps!

5 0
4 years ago
Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac
GalinKa [24]
<h2>a) The rate at which NO_2 is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen O_2 is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of NO = -\frac{1d[NO]}{2dt} = 0.066 M/s

Rate in terms of disappearance of O_2 = -\frac{1d[O_2]}{dt}

Rate in terms of appearance of NO_2= \frac{1d[NO_2]}{2dt}

1. The rate of formation of NO_2

-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}

\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s

2. The rate of disappearance of O_2

-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}

-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s

Learn more about rate law

brainly.com/question/13019661

https://brainly.in/question/1297322

7 0
3 years ago
The illustration above depicts a spring being compressed and then released. Changes in both potential and kinetic energy occur
Anestetic [448]

Answer:

C)

the potential energy is increasing through steps A & B, then decreases at C.

Explanation:

5 0
3 years ago
Read 2 more answers
5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points
labwork [276]

Answer: (A) H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)

(B) H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of H_{3}O^{+} in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

       Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)

 NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)

Hence, net ionic equation will be as follows.

       H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)

(B)  When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

     NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)

The net ionic equation is as follows.

        H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)

7 0
3 years ago
What’s the compound name for V2(S2O3)3
ohaa [14]

Answer: I HOPE THIS HELPS, HAVE A GREAT EARLY HALLOWEEN

Explanation:

PubChem CID: 1084

Molecular Formula: S2O3(2−) or O3S2-2

Synonyms: Thiosulphate THIOSULFATE ION sulfurothioate UNII-LLT6XV39PY Thiosulfate (S2O32-) More...

Molecular Weight: 112.13 g/mol

7 0
3 years ago
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