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3 years ago

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Parallelogram PQRS is shown below. Parallelogram P Q R S. Angle P is 116 degrees. Angle R is opposite to angle P. What are the m

easures of the three remaining angles? Measure of angle Q = 116 degrees, Measure of angle R = 64 degrees, Measure of angle S = 64 degrees Measure of angle Q = 116 degrees, Measure of angle R = 116 degrees, Measure of angle S = 116 degrees Measure of angle Q = 64 degrees, Measure of angle R = 64 degrees, Measure of angle S = 64 degrees Measure of angle Q = 64 degrees, Measure of angle R = 116 degrees, Measure of angle S = 64 degrees

1 answer:

aliya0001 [1]3 years ago

5 0

Answer:

64 degrees

Explanation:

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What is the value for the reaction: N2(g) + 2 O2(g) --> N2O4(g) in terms of K values from the reactions:

Valentin [98]

Answer : The correct expression will be:

$K=(K_1)^2\times K_2$

Explanation :

The chemical reactions are :

(1) $\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\rightleftharpoons NO(g)$ $K_1$

(2) $2NO(g)+O_2(g)\rightleftharpoons N_2O_4(g)$ $K_2$

The final chemical reaction is :

$N_2(g)+2O_2(g)\rightleftharpoons N_2O_4$ $K=?$

Now we have to calculate the value of $K$ for the final reaction.

Now equation 1 is multiply by 2 and then add both the reaction we get the value of 'K'.

If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.

If the two equations are added then equilibrium constant will be multiplied.

Thus, the value of 'K' will be:

$K=(K_1)^2\times K_2$

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3 years ago

How do we seperate a mixture of water and sugar

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Explanation:

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3 years ago

Artemon [7]

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Explanation:

using the equation :

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2 years ago

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

<u>Part 3 :</u>

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

<u>Part 4 :</u>

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago