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3 years ago

calculate the number of grams of magnesium in 50g of magnesium sulfide. Use percentage composition and gravimetric factor.

Chemistry

1 answer:

faltersainse [42]3 years ago

3 0

Answer:

50 gram of MgS contains =21.56 gram of Mg

Explanation:

Molecular mass of MgS is 56.37 g/mol.

Atomic mass of 1 mole Mg is 24.305

Therefore 56.37 gram of MgS contains 24.305 gram of Mg

1 gram of MgS contains $\frac{24.305}{56.37}$ gram of Mg

50 gram of MgS contains $\frac{24.305}{56.37}$ ×50 gram of Mg

=21.56 gram of Mg

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How much Oxygen would be required to react with 101 g of Methane?

AlexFokin [52]

Answer:

404g

Explanation:

6 0

3 years ago

Bromine (Br2) is produced by reacting HBr with O2, with water as a byproduct. The O2 is part of an air (21 mol % O2, 79 mol % N2

Karolina [17]

Answer:

The mole fractions:

$x_{HBr}=\frac{100mol}{318.5}=0.314$

$x_{Br_2}=\frac{78mol}{318.5}=0.245$

$x_{H_2O}=\frac{78mol}{318.5}=0.245$

$x_{O_2}=\frac{62.5mol}{318.5}=0.196$

Explanation:

The reaction described is:

$2 HBr (g) + 1/2 O_2 (g) \longrightarrow Br_2 (g) + H_2O (g)$

The limiting reactant is the HBr (oxygen is in excess).

a) The mass (in moles) balance for this sistem:

$n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *n_{HBr}*0.78$

(the 0.78 is because of the fractional conversion)

$n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *n_{HBr}*1.25$

(the 1.25 is because of the oxygen excess)

$n_{H_2O}=\frac{ 1 mol H_2O}{1 mol Br_2} *n_{Br_2}$

There is only one degree of freedom in this sistem, you can either deffine the moles of HBr you have or the moles of Br2 you want to produce. The other variables are all linked by the equations above.

b) Base of calculation 100 mol of HBr:

$nn_{HBr}=100 mol HBr$

$n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *100mol HBr*0.78$

$n_{Br_2}=78 mol Br2$

$n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *100 mol HBr*1.25$

$n_{O_2}=62.5 mol O_2$

$n_{H_2O}=n_{Br_2}= 78 mol$

$n_{total}=(78+78+100+62.5)mol= 318.5mol$

The mole fractions:

$x_{HBr}=\frac{100mol}{318.5}=0.314$

$x_{Br_2}=\frac{78mol}{318.5}=0.245$

$x_{H_2O}=\frac{78mol}{318.5}=0.245$

$x_{O_2}=\frac{62.5mol}{318.5}=0.196$

4 0

3 years ago

Complete the equation: monosaccharide monosaccharide → ______ water

Lady_Fox [76]

If I understand you correctly, then this is the equation for the breakdown of monosaccharide.
monosaccharide --> Carbon dioxide + water

5 0

3 years ago

A) week 1 b) week 2 c) week 3 d) week 4

nalin [4]

Try c I think is the most accurate one

4 0

3 years ago

Read 2 more answers

In the disproportionation reaction CI2 + H2Omc021-1.jpgHCIO + HCI, what describes the oxidation states of the substance Cl?

Sophie [7]

Reaction: CI2 + H2O ----> HCIO + HCI

Oxidations states:

The oxitation state of Cl2 = 0, because the oxidation state of an atom alone or a molucule with one kind of atom is always 0.

The oxidation state of Cl in HClO is +1 because the oxidation state of H is + 1, the oxidation state of O is - 2, and the molecule is neutral, so +1 + 1 - 2 = 0

The oxidation state of Cl in HCl is - 1, because the oxidation state of H is +1 and the molecule is neutral, so - 1 + 1 = 0.

Also, you shall remember that when an atom increases its oxidation state is is oxidized and when an atoms reduces its oxidations state it is reduced.

With that you conclude that the right option is the last statement: Cl has an oxidation number of 0 in Cl2. It is then reduced to CI- with an oxidation number of –1 in HCl and is oxidized to Cl+ with an oxidation number +1 in HClO.

8 0

4 years ago