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2 years ago

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Given a water solution that contains 1.704 kg of HNO:/kg H:O and has a specific gravity of 1.382 at 20 °C, express the compositi

on in the following ways: (a) Weight percent HNO (b) Pounds HNO3 per cubic foot of solution at 20 °c (c) Molarity (gmoles of HNOs per liter of solution at 20 °C)

1 answer:

TEA [102]2 years ago

4 0

Answer:

a) 63,0%

b) 54,4 Pounds HNO₃ per cubic foot of solution

c) 13,8 M

Explanation:

a) Weight percent is the ratio solute:solution times 100:

$\frac{1,704 kg HNO_3}{2,704 kg Solution}$ = <em>63,0%</em>

b) Pounds HNO₃ per cubic foot of solution at 20 °c

Pounds HNO₃:

1,704 kg $\frac{2,20462 pounds}{1 kg}$ = <em>3,7567 pounds</em>

Cubic foot:

2,704 kg $\frac{1 L}{1,382 kg}$ x $\frac{1 cubic foot }{28,3168 L}$ = <em>0,069 ft³</em>

Thus: $\frac{3,7567 pounds}{0,069ft^3} =$ = 54,4 Pounds HNO₃ per cubic foot of solution

c) Moles of HNO₃:

1704 g HNO₃ $\frac{1 mol HNO_3}{63,01 g }$ = <em>27,04 moles</em>

Liters of solution:

2,704 kg $\frac{1 L}{1,382 kg}$ = <em>1,96 L of solution</em>

Molarity:

$\frac{27,04 mol}{1,96 L}$ = 13,8 M

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1 year ago

At the beginning of the school year, a chalk company receives an order for 2000 boxes which is the largest order ever placed. Ea

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<u>259,000 g of chalk.</u>

Explanation:

<u>1) Data:</u>

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b) 175 g / box

c) % yield = 74%

<u>2) Formula: </u>

% yield = (theoretical yield / actual yield) × 100

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2 years ago

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:

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Explanation:

Given that:

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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

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

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

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

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:

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

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

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

$\Delta H^0 = -4466.1 \ kJ$

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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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2 years ago

A slice of pizza contains 29 g of carbohydrate, 13 g of protein and an unknown amount of fat. if the pizza contains 280 kcal, ho

Shalnov [3]

When different macronutrients (e.g., carbohydrates) are metabolized, they yield different amounts of energy, commonly measured in kilocalories.

They are given by the following conversion factors:

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If the slice of pizza contains 29 g carbohydrates, that means there are 4 kcal/g × 29 g = 116 kcal from carbohydrates. It contains 13 g protein, so there are 4 kcal/g <span>× 13 g = 52 kcal from protein.

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2 years ago

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