All categories

2 years ago

The temperature of 6.24 l of a gas is increased from 25.0°c to 55.0°c at constant pressure. the new volume of the gas is

1 answer:

6 0

At constant pressure, temperature is directly proportional to the volume and vise versa. The formula will be

V1/T1 = V2/T2

where V1 and T1 are the initial volume and temperature and V2 and T2 are the final volume and temperature. The temperature is in Kelvin and to convert Celsius to Kelvin add 273.

so, 6.24L/298 = V2/328
=6.87L

You might be interested in

Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)

gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}$ , where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr 7.96 torr 0 0 initial partial pressure

-x -x +x +x react

105.1-x 7.96-x x x equilibrium

Then:

$Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}$

$2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}$

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

$x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}$

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

8 0

2 years ago

Read 2 more answers

How many grams of calcium cyanide (Ca(CN)2) are contained in 0.79 mol of calcium cyanide?

Allushta [10]

I dont know but do you know da wae brudda?

3 0

2 years ago

List the number of each type of atom on the left side of the equation 2C6H14(l)+19O2(g)→12CO2(g)+14H2O(g) Enter your answers sep

murzikaleks [220]

Answer: The number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively

Explanation:

In a chemical equation, the chemical species are termed as reactants or products.

Reactants are defined as the species which react in the reaction and are written on the left side of the reaction arrow.

Products are defined as the species which are produced in the reaction and are written on the right side of the reaction arrow.

For the given chemical equation:

$2C_6H_{14}(l)+19O_2(g)\rightarrow 12CO_2(g)+14H_2O(g)$

On the reactant side:

Number of carbon atoms = (6 × 2) = 12

Number of hydrogen atoms = (14 × 2) = 28

Number of oxygen atoms = (2 × 19) = 38

Hence, the number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively

3 0

2 years ago

What is the result of adding KF to an equilibrium mixture of the weak acid HF?

blagie [28]

Answer D. Follow Le Chatelier's principle.

5 0

2 years ago

How many moles are in 8.30x10^23 molecules of H2o

LUCKY_DIMON [66]

You multiply avogadro's number to what you were given.
8.30x10^23 * 6. 0221409x10^23
=1.357*10^25

That should be the right answer but I'm not sure. It has been awhile since I have done this.

4 0

2 years ago

Read 2 more answers