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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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when 6.0 mol of oxygen are confined in a 36L vessel at 196°c, the pressure is 8atm. what is the new pressure for oxygen expands
kykrilka [37]

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<u>Explanation:</u>

To calculate the new pressure of the gas, we use the equation given by Boyle's Law.

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Mathematically,

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

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We are given:

P_1=8atm\\V_1=36L\\P_2=?atm\\V_2=48L

Putting values in above equation, we get:

8atm\times 36L=P_2\times 48L\\\\P_2=6atm

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For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
Aneli [31]
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
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Average bond enthalpy = 594.9kJ/mol/2
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