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Zanzabum
3 years ago
6

If the temperature is zero outside today and it's going to be twice as cold tomorrow, how cold will it be?

Chemistry
1 answer:
makkiz [27]3 years ago
4 0
<span>If you are measuring in farenheight, twice as cold is not possible.(0/2=0) However, if you are using the Celsius scale, zero degrees is actually 273.15 degrees above absolute zero. So, twice as cold would be -136.575 degrees Celsius. Similarly, absolute zero in Fahrenheit is -459.67. </span>
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\boxed{\pmb{\color{gold}{\sf{2SO_{2}(g) + O_{2}(g)\dashrightarrow 2SO_{3}(g)}}}}

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3 years ago
Read 2 more answers
What mass, in grams, of sodium sulfate is needed to make 230.5 g of a 3.5 % (m/m) aqueous solution of sodium sulfate?
andreev551 [17]

The percent concentration of a solution can be calculated from; mass of solute /mass of solution * 100. The mass of the solute here is 8.1 g.

<h3>What is concentration?</h3>

The term concentration refers to the amount of solute presnt in a solution. There are many ways of expressing concentration such as molarity, molality and percentage.

Here;

mass of solution = 230.5 g

Percent of solute =  3.5 %

3.5 = x/ 230.5 * 100

3.5 = 100x/230.5

230.5(3.5) = 100x

x = 230.5(3.5) /100

x = 8.1 g

Learn more about percent concentration: brainly.com/question/202460?

7 0
2 years ago
In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
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