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polet [3.4K]
3 years ago
8

How will you prepare hydrogen from zinc and dilute acid.​

Chemistry
1 answer:
oksian1 [2.3K]3 years ago
7 0
Hydrogen is basically prepared by action of dilute acids On metals.
Dilute Hydro chloric acid is added To zinc granules. The hydrogen produced is then passed through anhydrous calcium chloride or concentrated sulphuric acid. The hydrogen collected by upward delivery because it is lighter than air. If the reaction is too slow Copper (II) Sulphate solution can be added to speed up the solution.
Zn (s)+ 2HCl (aq)———ZnCl2 (aq)+ H2(g).
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Describe the term molarity as it relates to an acid or base.
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Read 2 more answers
Unit: Chemical Quantities
Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 1.993\,moles

2) The number of atoms of helium is:

x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)

x = 9.033\times 10^{23}\,atoms

3) The quantity of moles of carbon monoxide is:

n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.689\,moles

4) The number of molecules of sulfur dioxide is:

x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.355\times 10^{24}\,molecules

5) The quantity of moles of sodium chloride is:

n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.399\,moles

6) The number of formula units of magnesium iodide is:

x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)

x = 1.084\times 10^{24}\,f.u.

7) The quantity of moles of potassium permanganate is:

n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }

n = 1.214\,moles

8) The number of molecules of carbon tetrachloride is:

x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.506\times 10^{23}\,molecules

9) The quantity of moles of aluminium is:

n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 0.609\,moles

10) The number of molecules of oxygen difluoride is:

x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 2.120\times 10^{24}\,molecules

3 0
4 years ago
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