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3 years ago

5

A metal coin has certain properties that can be measured if the coin were to go from earth to moon the coins will remain the sam

e

1 answer:

PtichkaEL [24]3 years ago

6 0

Answer:

its matter

Explanation:

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Diano4ka-milaya [45]

Answer:

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5 0

3 years ago

The equilibrium of 2H 2 O(g) 2H 2 (g) + O 2 (g) at 2,000 K has a Keq value of 5.31 x 10-10. What is the Keq expression for this

nevsk [136]

Answer:

$5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

Explanation:

For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.

For a hypothetical reaction:

xA + yB ⇄ zC

The equilibrium constant is :

$Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }$

The given reaction involves the decomposition of H2O into H2 and O2

$2H_{2}O\rightleftharpoons 2H_{2} + O_{2}$

The equilibrium constant is expressed as :

$Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

Since Keq = 5.31*10^-10

$5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

3 0

3 years ago

Read 2 more answers

Where is most of the mass found in an atom

IgorLugansk [536]

massive livand that sarah or someone is how u do it

7 0

3 years ago

Read 2 more answers

Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc

Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is $C_5H_7N$

(b) The empirical formula for the given compound is $C_4H_5N_2O$

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = $\frac{6.17}{1.23}=5.01\approx 5$

For Hydrogen = $\frac{8.70}{1.23}=7.07\approx 7$

For Nitrogen = $\frac{1.23}{1.23}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is $C_5H_7N_1=C_5H_7N$

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = $\frac{4.12}{1.03}=4$

For Hydrogen = $\frac{5.19}{1.03}=5.03\approx 5$

For Nitrogen = $\frac{2.06}{1.03}=2$

For Nitrogen = $\frac{1.03}{1.03}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is $C_4H_5N_2O_1=C_4H_5N_2O$

6 0

3 years ago

The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c

crimeas [40]

Answer : The rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $785.0K$ = ?

$Ea$ = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $785.0K$

Now put all the given values in this formula, we get:

$\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]$

$K_2=1.45\times 10^{-2}s^{-1}$

Therefore, the rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

7 0

3 years ago