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NARA [144]
3 years ago
10

Determine the number of grams of magnesium oxide if you have 2.63 moles of magnesium oxide.

Chemistry
1 answer:
ANTONII [103]3 years ago
3 0

The number of grams of Magnesium oxide= 106 g

<h3>Further explanation</h3>

Given

2.63 moles of Magnesium oxide

Required

mass of Magnesium oxide

Solution

The mole is the number of particles contained in a substance

<em> 1 mol = 6.02.10²³ </em>

Moles can also be determined from the amount of substance mass and its molar mass

Mol = mass : Molar mass

Mass of Magnesium oxide(MgO, MW=40.304 g/mol) :

= mol x MW

= 2.63 x 40.304

= 106 g

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Describe, in detail, how to make 5.5 liters of a 2.5 molar HCl solution from a 10.5 molar HCl stock solution.
Misha Larkins [42]
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4 years ago
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Answer:

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5 0
3 years ago
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
Leokris [45]

Answer:

Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:

element & mass %

phosphorus & 39.18%

sulfur & 60.82%

Write the molecular formula of X.

Explanation:

The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.

Empirical formula calculation:                      

element:              phosphorus                       sulfur

co9mposition:      39.185%                            60.82%

divide with

atomic mass:          39.185/31.0 g/mol           60.82/32.0g/mol

                              =1.26mol                           1.90mol

smallest mole ratio:   1.26mol/1.26mol =1      1.90mol/1.26 mol =1.50

multiply with 2:          2                                         3

Hence, the empirical formula is:

P2S3.

Mass of empirical formula is:

158.0g/mol

Given, molecule has molar mass --- 316.25 g/mol

Hence, the ratio is:

316.25g/mol/158.0 =2

Hence, the molecular formula of the compound is :

2 x (P2S3)

=P_4S_6

8 0
3 years ago
The volume of 0.05 M H2SO4 is needed to completely neutralise 15ml of 0.1 M NaOH solution is
Feliz [49]
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)

V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)

V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}

V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL
5 0
3 years ago
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