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2 years ago

14

If a gas is initially at a pressure of nine ATM and a volume at 21 L at a temperature of 253K and the pressure is raise to 15 AT

M and the temperature is raised to 302K what will be the resulting volume of the gas

1 answer:

Tom [10]2 years ago

3 0

Answer:

15.0L

Explanation:

p/v = constan

(9*21)/253 =(15v)/ 302

v = (9*21*302)/(15*253)

v=15.0

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Someone please help i don’t have much time left

REY [17]

Answer: Energy of reactants = 30, Energy of products = 10

Exothermic

Activation energy for forward reaction is 10.

Explanation:

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and $\Delta H$ for the reaction comes out to be negative.

Energy of reactants = 30

Energy of products = 10

Thus as energy of the product < energy of the reactant, the reaction is exothermic.

Activation energy $(E_a)$ is the extra energy that must be supplied to reactants in order to cross the energy barrier and thus convert to products.

$E_a$ for forward reaction is (40-30) = 10.

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2 years ago

A cylinder was charged with 1.25 atm of oxygen gas, 6.73 atm of argon, and 0.895 atm of xenon. What is the mole fraction of each

katrin2010 [14]

Considering the Dalton's partial pressure, the mole fraction of each gas is:

$x_{oxygen}$ = 0.14
$x_{argon}$ = 0.76
$x_{xenon}$ = 0.10

<h3>Dalton's partial pressure</h3>

The pressure exerted by a particular gas in a mixture is known as its partial pressure.

So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

$P_{T} =P_{1} +P_{2} +...+P_{n}$

where n is the amount of gases in the gas mixture.

This relationship is due to the assumption that there are no attractive forces between the gases.

Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of a component to the number of moles of all the components present.

So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:

$P_{A} =x_{A} P_{T}$

In summary, the total pressure in a mixture of gases is equal to the sum of partial pressures of each gas.

Mole fraction of each gas

In this case, you know that:

$P_{oxygen }$ = 1.25 atm
$P_{argon}$ = 6.73 atm
$P_{xenon}$ = 0.895 atm
$P_{T} =P_{oxygen} +P_{argon}+P_{xenon}$ = 1.25 atm + 6.73 atm + 0.895 atm= 8.875 atm

Then:

$P_{oxygen} =x_{oxygen} P_{T}$
$P_{argon} =x_{argon} P_{T}$
$P_{xenon} =x_{xenon} P_{T}$

Substituting the corresponding values:

1.25 atm= $x_{oxygen}$ 8.875 atm
6.73 atm= $x_{argon}$ 8.875 atm
0.895 atm= $x_{xenon}$ 8.875 atm

Solving:

$x_{oxygen}$ = 1.25 atm÷ 8.875 atm= 0.14
$x_{argon}$ = 6.73 atm÷ 8.875 atm= 0.76
$x_{xenon}$ = 0.895 atm÷ 8.875 atm=0.10

In summary, the mole fraction of each gas is:

$x_{oxygen}$ = 0.14
$x_{argon}$ = 0.76
$x_{xenon}$ = 0.10

Learn more about Dalton's partial pressure:

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1 year ago

Given the balanced equation, 2 H2 + O2 -> 2 H2O, answer the following

9966 [12]

Answer: $O_2$ is the limiting reagent

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} H_2=\frac{10.0g}{2g/mol}=5.00moles$

$\text{Moles of} O_2=\frac{20.0g}{32g/mol}=0.625moles$

The balanced chemical reaction is :

$2H_2+O_2\rightarrow 2H_2O$

According to stoichiometry :

1 moles of $O_2$ require = 2 moles of $H_2$

Thus 0.625 moles of $O_2$ will require= $\frac{2}{1}\times 0.625=1.25moles$ of $H_2$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent as it is present more than the required amount.

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What are all the stages of mitosis? (please make it long)

soldier1979 [14.2K]

Mitosis is conventionally divided into 5 phases, which include prophase, prometaphase, metaphase, anaphase and telophase and cytokinesis.

Interphase

Before coming into mitosis, a mobile spends a length of its increase underneath interphase.

Prophase

Prophase straight away follows the S and G2 levels of the cycle and is marked by way of condensation of the genetic fabric to form compact mitotic chromosomes composed of chromatids attached at the centromere.

Prometaphase

In the prometaphase, the nuclear envelop disintegrates. Now the microtubules are allowed to extend from the centromere to the chromosome.

Metaphase

At this level, the microtubules start pulling the chromosomes with equal pressure and the chromosome ends up in the center of the cell. This area is referred to as the metaphase plate.

Anaphase

The splitting of the sister chromatids marks the onset of anaphase. These sister chromatids end up the chromosome of the daughter nuclei.

Telophase

The chromosomes that cluster at the two poles start coalescing into an undifferentiated mass, because the nuclear envelope begins forming round it.

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