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zaharov [31]
3 years ago
5

How many miles of MgCI2 are there in 345 g of the compound?

Chemistry
1 answer:
Valentin [98]3 years ago
8 0
MgCl2
Mg=24
Cl=35.5×2 = 71
rmm= 71+24=95g
1 mole = molar mass
1 mole = 95g
1g = 1/95 moles
345g= 1/95 × 345
3.6 moles
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When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa
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Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

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q_{boiling} = mc \Delta T

where

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q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18  \ J/g^0 C \times (100 - 25)^0 C

q_{boiling} = 10.9725 \times 10^6 \ J

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

q_{combustion} =-  \dfrac{q_{boiling}}{0.15}

q_{combustion} =-  \dfrac{10.9725 \times 10^6 J}{0.15}

q_{combustion} = -7.315 \times 10^{7} \ J

q_{combustion} = -7.315 \times 10^{4} \ kJ

For heptane; the equation for its combustion reaction can be written as:

C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}

The standard enthalpies of the  products and the reactants are:

\Delta H _f   \ CO_{2(g)} = -393.5 kJ/mol

\Delta H _f   \ H_2O_{(g)} = -242 kJ/mol

\Delta H _f   \ C_7H_{16 }_{(g)} = -224.4 kJ/mol

\Delta H _f   \ O_{2{(g)}} = 0 kJ/mol

Therefore; the standard enthalpy for this combustion reaction is:

\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}

\Delta H^0 =( 7  \ mol ( -393.5 \ kJ/mol)  + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11  \ mol  (0 \ kJ/mol))

\Delta H^0 = (-2754.5 \ \  kJ -  1936 \ \  kJ+224.4 \  \ kJ+0 \ \  kJ)

\Delta H^0 = -4466.1 \ kJ

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn 7.315 \times 10^{4} \ kJ heat released is:

n_{fuel} = \dfrac{q}{\Delta \ H^0}

n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1  \ kJ}

n_{fuel} = 16.38  \ mol \ of \ C_7 H_{16

Since number of moles = mass/molar mass

The  mass of the fuel is:

m_{fuel } = 16.38 mol \times 100.198 \ g/mol}

m_{fuel } = 1.641 \times 10^{3} \ g

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = \dfrac{1.641 \times 10^3 \ g }{0.78  g/mL} \times \dfrac{L}{10^3 \ mL}

volume of the fuel  = 2.104 L fuel

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