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Flura [38]
2 years ago
14

What volume of gold would be equal in mass to a piece of copper with a volume of 141 ml? the density of gold is 19.3 g/ml; the d

ensity of copper is 8.96 g/ml?
Chemistry
1 answer:
sergejj [24]2 years ago
8 0
We are tasked to solve for the volume of Gold given that the mass of Gold is equivalent to that of a copper.

Details:

Density of Copper= 8.96 g/ml
Volume of Copper=141 ml
mass of Gold = mass of Copper
Density of Gold=19.3 g/ml

In order to solve for the mass of copper, we need to use the density formula
Density= mass/volume

Since we are to solve for the mass of copper
mass of copper= Density of Copper * Volume of Copper
mass of copper= 8.96 g/ml* 141 ml= 1263.36 g
Thus,
mass of gold=mass of copper=1263.36 g
Hence,
Using still the density formula to solve for the volume of gold,
Volume of gold=mass of gold/ Density of gold
Volume of gold=1263.36 g/ 19.3 g/ml = 65.46 mL

Therefore, the volume of the gold must be 65.46 mL in order to have the same mass of a copper
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Answer:

The molar mass of the compound is:- 168.82 g/mol

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Explanation:

(a)

Using ideal gas equation as:

PV=nRT

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R is Gas constant having value = 0.0821 L.atm/K.mol

Also,  

Moles = mass (m) / Molar mass (M)

Density (d)  = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

PM=dRt

Given that:-

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The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = \frac {1}{101325} P (atm)

20000 Pa = \frac {20000}{101325} atm

Pressure = 0.1974 atm

Temperature = 330 K

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(b)

Given that:

Pressure = 152 Torr

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Using ideal gas equation as:

PV=nRT

R = 62.3637\text{torr}mol^{-1}K^{-1}

Applying the equation as:

152 Torr × 0.25 L = n × 62.3637 L.torr/K.mol × 298 K

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2 years ago
Give the product for the reaction of 1-butene with methanol in the presence of acid. The mechanism is the same as the mechanism
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Answer:

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<u />

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1-butanol will be formed when water is added to 1-butene. The mechanism has the same principle but not the same product.

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Explanation:

Hope this answer helps.

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