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ollegr [7]
3 years ago
6

Write an equation between the carbonate and the acid​

Chemistry
1 answer:
allochka39001 [22]3 years ago
8 0

Answer:

In general an acid reacts with a carbonate or hydrogen-carbonate to produce a salt, carbon dioxide gas and water.

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False everything involves matter

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A cell divides to form two cells during a process called
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I believe its called mitosis. Mitosis is a type of cell division that results in two daughter cells each having the same number and kind of chromosomes as the parent nucleus, typical of ordinary tissue growth. Hope this helps:) 
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0.58 mol of Mg contains how many atoms? please show work
Jet001 [13]

The number of atoms present in 0.58 mole of magnesium, Mg is 3.49×10²³ atoms

<h3>Avogadro's hypothesis </h3>

1 mole of Mg = 6.02×10²³ atoms

<h3>How to determine the atoms in 0.58 mole of Mg </h3>

1 mole of Mg = 6.02×10²³ atoms

Therefore,

0.58 mole of Mg = 0.58 × 6.02×10²³

0.58 mole of Mg = 3.49×10²³ atoms

Thus, 3.49×10²³ atoms are present in 0.58 mole of Mg

Learn more about Avogadro's number:

brainly.com/question/26141731

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3 0
2 years ago
Determine the number of protons, neutrons and electrons in each of the following:
Kisachek [45]
Cu+ p=29 e=28 n=34
S2- p=16 e=18 n=16
Pb4+ p=82 e=78 p=125

I hope i did it right :))
5 0
3 years ago
A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so
omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s).

  • Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
  • Secondly, addition of liquid ammonia would form a precipitate with silver: AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq); Silver hydroxide at higher temperatures decomposes into black silver oxide: 2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g).
  • Thirdly, we also know we have Hg_2^{2+} in the mixture, since addition of potassium chromate produces a yellow precipitate: Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s). The latter precipitate is yellow.
3 0
4 years ago
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