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Ainat [17]
3 years ago
12

List the number of each type of atom on the right side of the equation hbr(aq)+2naoh(aq)→2nabr(s)+h2o(l)

Chemistry
2 answers:
ser-zykov [4K]3 years ago
5 0

On the basis of the given unbalanced equation, that is:

HBr (aq) + 2NaOH (aq) → 2NaBr (s) + H2O (l)

On the right side of the equation, there are 2 atoms of sodium (Na), 2 atoms of bromine (Br), 2 atoms of hydrogen (H), and 1 atom of oxygen (O₂).

After balancing the equation correctly we get:

HBr (aq) + NaOH (aq) → NaBr (s) + H2O (l)

On the right side, one atom of Na, 1 atom of Br, 1 atom of H and one atom of O₂.

AlexFokin [52]3 years ago
4 0

Answer:

As per the given unbalanced equation:-

HBr(aq) + 2NaOH(aq) --> 2NaBr(s) + H2O(l)

Na:-2

Br:-2

H:- 2

O:- 1

If correctly balanced then

HBr(aq) + NaOH(aq) --> NaBr(s) + H2O(l)

Na:-1

Br:-1

H:- 2

O:- 1

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Answer:

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2 years ago
Elements ending in the electron configurations ns^1 are highly reactive metals . What family does these elements belong to ?
myrzilka [38]

Answer:

<h3>A . Alkali metals</h3>

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6 0
3 years ago
A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

8 0
3 years ago
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8 0
3 years ago
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