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2 years ago

Question 1

Chemistry

1 answer:

Bumek [7]2 years ago

7 0

Answer:

the average kinetic energy will double as well

Explanation:

According to the kinetic molecular theory of gases, the average kinetic energy of the particles in a substance is directly proportional to the absolute temperature of the gas (in Kelvin).

Mathematically:

$KE=\frac{3}{2}kT$

where

KE is the average kinetic energy of the particles in a gas

k is the Boltzmann constant

T is the absolute temperature of the gas

From the equation, we see that

$KE\propto T$

In this problem, the absolute temperature of the gas is doubled, so the new temperature is

$T'=2T$

Therefore, the new kinetic energy of the molecules will be

$KE'\propto T' = 2T \propto 2KE$

So, the average kinetic energy will double as well.

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The Keq for the equilibrium below is 5.4 × 1013 at 480.0 °C. 2NO (g) + O2 (g) 2NO2 (g) What is the value of Keq at this temperat

kodGreya [7K]

Answer: The equilibrium constant for $NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g)$ equation is $1.36\times 10^{-7}$

Explanation:

The given chemical equation follows:

$2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)$

The value of equilibrium constant for the above equation is $K_{eq}=5.4\times 10^{13}$

Calculating the equilibrium constant for the given equation:

$NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g)$

The value of equilibrium constant for the above equation will be:

$K'_{eq}=\frac{1}{\sqrt{K_{eq}}}\\\\K'_{eq}=\frac{1}{\sqrt{5.4\times 10^{13}}}\\\\K'_{eq}=1.36\times 10^{-7}$

Hence, the equilibrium constant for $NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g)$ equation is $1.36\times 10^{-7}$

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1 year ago

the half-life of a certain radioactive element is 1250 years. what percent of the atom remains after 7500 years?

Olenka [21]

Maybe 24% not sure try researching it on google

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2 years ago

An atom of the element zinc has an atomic number of 30 and a mass number of 65. How many protons does an uncharged zinc atom hav

navik [9.2K]

Answer : The correct option is, 30 protons

Explanation :

Element = Zinc

Atomic number = 30

Atomic mass number = 65

As we know that the atomic number is equal to the number of electrons and number of protons.

Atomic number = Number of electrons = Number of protons = 30

Number of neutrons = Atomic mass - Number of protons = 65 - 30 = 35

Therefore, the number of protons an uncharged zinc atom have 30 protons.

6 0

2 years ago

Read 2 more answers

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

2 years ago

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

2 years ago