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iVinArrow [24]
2 years ago
11

Which type of changes are melting, freezing, and boiling?​

Chemistry
2 answers:
sukhopar [10]2 years ago
8 0

Answer:

the changes are physical changes

tatiyna2 years ago
4 0

Answer:

Adding heat.

Decreasing and increasing temperatures.

Explanation:

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What are the correct coefficients when this equation is balanced?
larisa86 [58]

4, 3, 1

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O2 comes in pairs, so you only need <u>3</u> of those to get six oxygens.

And 1 is the place holder for the right side since we got the left to match perfectly.

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Here, acceleration due to gravity(a) is assumed as 10m/s².We can also take it as 9.8m/s²

Explanation:Brick) Weight=m * a=1 * 10=10N\\Bag of books) Mass=f/a=25/10=2.5kg\\Bag of potatoes) Weight=m * a=10 * 10=100N\\Bag of cement) Mass=f/a=400/10=40kg

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The smell of hot food moves faster than the smell of cold food. Which property of gas is responsible for this?
miskamm [114]
When pressure is added to a gas the molecules bounce around really fast and push against the walls of its container. That's why when you squeeze an empty water bottle you can crush it all the way. The smaller the container the more the molecules hit the walls faster and that's what creates pressure. While moving around the molecules get hot as they speed up. Heat = Pressure. 
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3 years ago
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sesenic [268]
Lol your answer is c
3 0
3 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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