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Strike441 [17]
2 years ago
15

How many grams of potassium chloride is produced with 2.5g of pure potassium

Chemistry
1 answer:
vazorg [7]2 years ago
3 0

Answer:

The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want.

change 25 grams of potassium chlorate to moles.

calculate the gram molecular mass of potassium chlorate.

Chlorate is Cl with 3 oxygens. ate = saturated. Chlorine has seven valance electrons when it is saturated six of these electrons are used by oxygen ( 2 electrons per oxygen) leaving only 1 electron.

1 K x 39 grams/mole

+1 Cl x 35.4 grams/ mole

+3 O x 16 grams/ mole

= 122.4 grams / mole Potassium Chlorate

25

122.4

= moles.

2.05 moles of Potassium Chlorate.

There is a 1:1 mole ratio. 1 mole of Potassium Chlorate will produce 1 mole of Potassium Chloride.

2.05 moles of Potassium Chlorate will produce 2.05 moles of Potassium Chloride.

Find the gram molecular mass of Potassium Chloride.

1 K x 39 = 39

+1 Cl x 35.4 = 35.4

= 74.4 grams / mole.

2.05 moles x 74.4 grams/ mole = 15.2 grams

Explanation:

big brain ;)

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How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with
konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = a=31 cm^2

Thickness of the gold plating  = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3

Density of the gold = d=19.3 g/cm^3

m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g

Moles of gold = \frac{29.915 g}{197 g/mol}=0.152 mol

Au^{3+}+3e^-\rightarrow Au

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

\frac{3}{1}\times 0.152 mol=0.456 mol of electrons

Number of electrons = N =0.456\times \times 6.022\times 10^{23}

Charge on single electron = q=1.6\times 10^{-19} C

Total charge required = Q

Q=N\times q

Amount of current passes = I = 8 Ampere

Duration of time  = T

I=\frac{Q}{T}

T=\frac{N\times q}{I}

=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0
3 years ago
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