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V125BC [204]
3 years ago
8

Mercury has a specific gravity of 13.6. how many milliliters of mercury have a mass of 0.35 kg

Chemistry
1 answer:
SashulF [63]3 years ago
6 0

The formula for specific gravity is:

Specific gravity = \frac{\rho _{substance}}{\rho _{water}}

where \rho _{substance} is the density of the substance and \rho _{water} is the density of water.

The density of water, \rho _{water} = 1 g/mL

Substituting the values in above formula we get,

13.6 = \frac{\rho _{substance}}{1}

\rho _{substance} = 13.6 g/mL

The formula of density is:

density = \frac{mass}{volume}

The density of mercury is 13.6 g/mL

The mass of mercury is 0.35 kg = 0.35 kg \times 1000 \frac{g}{kg} = 350 g

Substituting the values in density formula:

13.6 g/mL = \frac{350 g}{volume}

volume = \frac{350 g}{13.6 g/mL} = 25.73 mL

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Determine how many liters of hydrogen adjusted to STP there are in a 50.0 liter steel cylinder if the pressure inside is 100.0 a
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Answer : The volume of hydrogen gas at STP is 4550 L.

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 100.0 atm

P_2 = final pressure of gas at STP = 1 atm

V_1 = initial volume of gas = 50.0 L

V_2 = final volume of gas at STP = ?

T_1 = initial temperature of gas = 27.0^oC=273+27.0=300K

T_2 = final temperature of gas at STP = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{100.0atm\times 50.0L}{300K}=\frac{1atm\times V_2}{273K}

V_2=4550L

Therefore, the volume of hydrogen gas at STP is 4550 L.

3 0
3 years ago
The mole ratio of two reactants in a chemical equation is determined by
Ipatiy [6.2K]

Answer:

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How many grams are in 3.14 x 1015 molecules of CO?
Vesnalui [34]

Answer:

Explanation:

Not Many

1 mol of CO has a mass of

C = 12

O = 16

1 mol = 28 grams.

1 mol of molecules = 6.02 * 10^23

x mol of molecules = 3.14 * 10^15        Cross multiply

6.02*10^23 x = 1 * 3.14 * 10^15             Divide by 6.02*10^23

x = 3.14*10^15 / 6.02*10^23

x = 0.000000005 mols

x = 5*10^-9

1 mol of CO has a mass of 28

5*10^-9 mol of CO has a mass of x                        Cross Multiply

x = 5 * 10^-9 * 28

x = 1.46 * 10^-7 grams

Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample

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