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Aleksandr-060686 [28]
3 years ago
12

How do you find the amount of moles is .032 grams of water and whats the answer

Chemistry
1 answer:
masha68 [24]3 years ago
6 0

Answer:

\boxed {\boxed {\sf 0.0018 \ mol \ H_2 O }}

Explanation:

First, we need to find the molecular mass of water (H₂O).

H₂O has:

  • 2 Hydrogen atoms (subscript of 2)
  • 1 Oxygen atom (implied subscript of 1)

Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.

  • Hydrogen: 1.0079 g/mol
  • Oxygen: 15.9994 g/mol

There are 2 hydrogen atoms, so multiply the mass by 2.

  • 2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol

Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.

  • 2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol

Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

0.032 \ g  \ H_2 O* \frac{ 1 \ mol \ H_2 O}{18.0152 \ g \ H_2 O}

Multiply. The grams of H₂O will cancel out.

0.032 * \frac{1 \ mol \ H_2 O}{18.0152 }

\frac{0.032 *1 \ mol \ H_2 O}{18.0152 }

0.00177627781 \ mol \ H_2 O

The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.

The 7 in the hundred thousandth place tells us to round up.

0.0018 \ mol \ H_2 O

There are about <u>0.0018 moles in 0.032 grams.</u>

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3 0
3 years ago
Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
What is the change in boiling point for a 0.615m solution of Mgl2 in water?
ivanzaharov [21]

Answer :  The change in boiling point is, 0.94^oC

Explanation :

Formula used :

\Delta T_b=i\times K_f\times m

where,

\Delta T_b = change in boiling point = ?

i = Van't Hoff factor = 3 (for MgI₂ electrolyte)

K_f = boiling point constant for water = 0.51^oC/m

m = molality  = 0.615 m

Now put all the given values in this formula, we get

\Delta T_b=3\times (0.51^oC/m)\times 0.615m

\Delta T_b=0.94^oC

Therefore, the change in boiling point is, 0.94^oC

4 0
3 years ago
What is the maximum amount in moles of P2O5 that can theoretically be made from 235 g of P4 and excess oxygen?
Lady_Fox [76]
The correct answer would be 3.8mol
5 0
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