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Elanso [62]
3 years ago
10

Equation balancing

Chemistry
1 answer:
meriva3 years ago
4 0

<u>Answer:</u>

<u>For a:</u> The balanced equation is 2S(s)+3O_2(g)\rightarrow 2SO_3(g)

<u>For c:</u> The balanced equation is 2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

<u>Explanation:</u>

A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.

  • <u>For (a):</u>

The given unbalanced equation follows:

S(s)+O_2(g)\rightarrow SO_3(g)

To balance the equation, we must balance the atoms by adding 2 infront of both S(s) and SO_3 and 3 in front of O_2

For the balanced chemical equation:

2S(s)+3O_2(g)\rightarrow 2SO_3(g)

  • <u>For (b):</u>

The given balanced equation follows:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)

The given equation is already balanced.

  • <u>For (c):</u>

The given unbalanced equation follows:

2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+H_2O(l)

To balance the equation, we must balance the atoms by adding 2 infront of H_2O(l)

For the balanced chemical equation:

2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

  • <u>For (d):</u>

The given balanced equation follows:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

The given equation is already balanced.

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How many hydrogen atoms are in 5.20 mol of ammonium sulfide?
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2) That means that there are 2*4 = 8 atoms of hydrogen in each molecule of ammoium sulfide, so in 5.20 mol of molecules will be 8 * 5.20 mol = 41.6 moles of atoms of hydrogen

3) To pass to number of atoms multiply by Avogadro's number: 6.022 * 10^23

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A solution is made by adding 35.5 mL of concentrated hydrochloric acid ( 37.3 wt% , density 1.19 g/mL1.19 g/mL ) to some water i
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Answer:

1.73 M

Explanation:

We must first obtain the concentration of the concentrated acid from the formula;

Co= 10pd/M

Where

Co= concentration of concentrated acid = (the unknown)

p= percentage concentration of concentrated acid= 37.3%

d= density of concentrated acid = 1.19 g/ml

M= Molar mass of the anhydrous acid

Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1

Substituting values;

Co= 10 × 37.3 × 1.19/36.5

Co= 443.87/36.6

Co= 12.16 M

We can now use the dilution formula

CoVo= CdVd

Where;

Co= concentration of concentrated acid= 12.16 M

Vo= volume of concentrated acid = 35.5 ml

Cd= concentration of dilute acid =(the unknown)

Vd= volume of dilute acid = 250ml

Substituting values and making Cd the subject of the formula;

Cd= CoVo/Vd

Cd= 12.16 × 35.5/250

Cd= 1.73 M

7 0
3 years ago
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