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adelina 88 [10]
3 years ago
7

what is the mass of carbon dioxide which contain the same number of molecules as are contained in 14 gram of oxygen?​

Chemistry
1 answer:
Ratling [72]3 years ago
3 0

Answer:

Mass of CO2 WILL BE ~ 9.33 g

Explanation:

Moles of O2 = 14/18

Let the mass of CO2 be x

Then moles of CO2 will be = x/12

moles of CO2 = moles of O2

x/12 = 14/16

x = 9.33 grams

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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
Mole conversions.<br> Find the number of moles of argon in 607g of argon?
bogdanovich [222]

Answer:

24249.65 mol

Explanation:

n=MM × m

n= 39.95 ×607

n=24249.65

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Answer:

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it protects everything inside our body

Explanation:

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Answer:

A lower ph is always more acidic, due to the increased concentration of hydrogen ions in the solution/substance.

A ph of 3 is 100 times more acidic than a pH of 5, and this is due to the increments on the scale.

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