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3 years ago

Answer all parts of the following questions on Particle Stoichiometry.

1 answer:

3 0

<span>1. Translate, predict the products, and balance the equation above.

Li + Cu(NO3)2 = Li(NO3)2 + Cu

2. How many particles of lithium are needed to produce 125 g of copper?

125 g Cu ( 1 mol / 63.55 g ) (1 mol Li / 1 mol Cu ) ( 6.022 x 10^23 particles / 1 mol ) = 1.18x10^24 Li particles

3. How many grams of lithium nitrate are produced from 4.83E24 particles of copper (II) nitrate?

</span>4.83E24 particles of copper (II) nitrate ( 1 mol / 6.022x10^23 particles ) (1 mol Li(NO3)2 / 1 mol Cu(NO3)2 ) ( 130.95 g / 1 mol ) = 1043.77 grams Li(NO3)2

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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

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

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

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

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

3 0

4 years ago

What is the color of the mineral when it is in powdered form

MA_775_DIABLO [31]

Answer: streak

Explanation: Streak is the color of the powder of a mineral. To do a streak test, you scrape the mineral across an unglazed porcelain plate. The plate is harder than many minerals, causing the minerals to leave a streak of powder on the plate.

7 0

3 years ago

Oxygen has an atomic number of 8 and most commonly, a mass number of 16. thus, what is the atomic mass of an oxygen atom?

dexar [7]

Oxygen has atomic mass 16 amu

7 0

3 years ago

What effect does a high carbon level have on a deep ocean

oksian1 [2.3K]

Explanation:

High carbon concentration in the deep ocean means increased absorption of carbon to the atmosphere resulting to even greater and harmful amounts of carbon in the atmosphere. Therefore we need to keep a close eye of the deep ocean in the quest to monitor and pump out excess carbon from this part of marine life.

8 0

3 years ago

How do I solve this?

ira [324]

Sorry, I can't really see the question )-:

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4 years ago