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AleksandrR [38]
3 years ago
12

B) How many g of O2 are needed to produce 6.0 mol of H2O?

Chemistry
1 answer:
musickatia [10]3 years ago
8 0

Answer:

Mass of oxygen that will be required to produce 6  mole of water is 96g .

Explanation:

Mass of oxygen = ?

Atomic mass of oxygen is =16g

Molecular mass of O₂= 32 g

Now, according to question statement we need to calculate oxygen needed to produce 6 mol of water .

In 1 mol of water  ,there is O₂/2 is present or we can say that 1 oxygen atom is present .

We also know that ,

1 mol of water has = 1 mole of oxygen

Therefore ,

6 moles of water will have 6 moles of oxygen atom

Mass of 6 moles of oxygen atom can be calculated as :

Mole =given mass/atomic mass

6=X /16

X = 16 x 6

=96g

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Why is water essential to living things? (1 pc
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If you have 35 ml of gas in a closed container under 1.23 atm of pressure, what will the volume be if the pressure increased to
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Answer:

New volume is 14.35 mL

Explanation:

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.
azamat

Answer:

(a) m_{gold}=7.322g

(b)

V_{gold}=0.379cm^3

V_{copper}=0.122cm^3

(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3}  \\\\V_{copper}=0.122cm^3

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

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