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3 years ago

Calculate the enthalpy change when 4.00 mol Cl2O7 is produced according to the following balanced equation:

Chemistry

1 answer:

marissa [1.9K]3 years ago

8 0

So the equation is,

2Cl2(g) + 7O2(g) + 130kcal -> 2Cl2O7(g)

from this , ΔH°rxn= $130kcal.mol^{-1}$

$mol^{-1}$ is mentioned because it is for per mole of reaction. So for 4. moles of the product $2Cl_{2} O_{7}$ we need 4/2 moles of reaction to be used to calculate associated enthalphy change for the reaction.

Therefore $4/2*130=260 kcal$

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If you had a beaker full of black powder which of the following could you have?

Evgesh-ka [11]

A beaker of black powder can be a mixture with different substances (i.e gunpowder: a mixture of sulfur, charcoal, etc.), an element which can probably be carbon (known as black), and a compound as well which can be potassium nitrate. Hence, the answer is D.

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3 years ago

Read 2 more answers

Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) â†’ 2SO3(g) Suppose the volume of this system is c

oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

$2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}$

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be $P_1$

and the total pressure at the final equilibrium be $P_2$

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝ 1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

$P_1V_1 = P_2V_2$

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

$V_2 = \dfrac{V_1}{\dfrac{3}{2}} ----- (1)$

also;

Thus ;

$P_1V_1 = P_2( \dfrac{V_1}{\frac{3}{2}})$

$P_1V_1 = P_2 * 2 \dfrac{V_1}{3}$

$3 P_1 V_1 = 2 P_2 V_1$

Dividing both sides by $V_1$

$3P_1 = 2P_2$

$P_2 =P_1 \dfrac{3}{2} ----- (2)$

From ;

$P_1V_1 = P_2V_2$

$P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}$

$P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{2 }{3}}*V_1$

$P_2 V_2 = P_1 V_1$

Thus; The new equilibrium total pressure will be increased to one-half to initial total pressure.

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3 years ago

A sample originally contains 24 grams of a radioactive isotope. After 18 days, 18.8 grams of the isotope remain in the sample. W

LenKa [72]

Around 36 or 37 days, because of the decimal

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2 years ago

Which of the following examples illustrates a number that is correctly rounded to three significant figures?

Naily [24]

Answer:

c. 20.0332 g to 20,0 g

Explanation:

A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.

<em>Which of the following examples illustrates a number that is correctly rounded to three significant figures? </em>

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3 years ago

A compound contains 72% magnesium and 28% nitrogen. What is its empirical formula?

SashulF [63]

So the empirical formula is Mg3N2

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