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sergey [27]
3 years ago
7

Many common liquids have boiling points that are less than 110°C, whereas most metals are solids at room temperature and have mu

ch higher boiling points.
The boiling point of iodomethane, , is 42.4°C. What is the equivalent absolute temperature?

Temperature =
K

The boiling point of bismuth is 1833 K. What is the equivalent temperature on the Celsius scale?

Temperature =
°C
Chemistry
1 answer:
Semmy [17]3 years ago
7 0

Answer:

1. 315.4 K

2. 1560 °C

Explanation:

To convert from celsius to Kelvin, the following formula can be used:

T(K) = T(°C) + 273

Where:

T(K) => Temperature in Kelvin

T(°C) => Temperature in degree celsius

1. Determination of the temperature in Kelvin.

Temperature (T) in °C = 42.4 °C.

Temperature (T) in K =?

T(K) = T(°C) + 273

T(K) = 42.4 °C + 273

T(K) = 315.4 K

2. Determination of the temperature in degree Celsius.

Temperature (T) in K = 1833 K

Temperature (T) in °C =?

T(K) = T(°C) + 273

1833 = T(°C) + 273

Collect like terms

T(°C) = 1833 – 273

T(°C) = 1560 °C

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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
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<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

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