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3 years ago

The final temperature and pressure in a container is 234K at 210 kPa.

Chemistry

1 answer:

andre [41]3 years ago

5 0

Answer:

Initial pressure = 157 kpa (Approx)

Explanation:

Given:

final temperature = 234 K

final pressure = 210 kpa

Initial temperature = 175 K

Find:

Initial pressure

Computation:

Initial pressure / Initial temperature = final pressure / final temperature

Initial pressure / 175 = 210 / 234

Initial pressure = 157 kpa (Approx)

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What is “Polymerization” is your answer

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3 years ago

Sodium metal reacts with water to produce hydrogen gas.

Darina [25.2K]

Answer:

Explanation:

Question 6 options:

A single replacement reaction takes place because sodium is less reactive than hydroxide ions.

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3 years ago

A certain substance X has a normal freezing point of 5.6 °C and a molal freezing point depression constant Kf-7.78 °C-kg·mol-1.

Harrizon [31]

Answer:

27.60 g urea

Explanation:

The freezing-point depression is expressed by the formula:

ΔT= Kf * m

In this case,

ΔT = 5.6 - (-0.9) = 6.5 °C

Kf = 7.78 °C kg·mol⁻¹

m is the molality of the urea solution in X (mol urea/kg of X)

First we calculate the molality:

6.5 °C = 7.78 °C kg·mol⁻¹ * m

m = 0.84 m

Now we calculate the moles of urea that were dissolved:

550 g X ⇒ 550 / 1000 = 0.550 kg X

0.84 m = mol Urea / 0.550 kg X

mol Urea = 0.46 mol

Finally we calculate the mass of urea, using its molecular weight:

0.46 mol * 60.06 g/mol = 27.60 g urea

7 0

3 years ago

For a principal quantum number n, how many atomic orbitals are possible?

slavikrds [6]

Answer:

The total number of orbitals for a given n value is n2.

Explanation:

For a hydrogen atom with n=1, the electron is in its ground state; if the electron is in the n=2 orbital, it is in an excited state.

5 0

3 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago