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makvit [3.9K]
2 years ago
10

Is sodium potassium nitrate exo or endo?​

Chemistry
2 answers:
ANEK [815]2 years ago
3 0

endothermic

Explanation:

I believe it's an endothermic but I'm not all too sure

mixas84 [53]2 years ago
3 0

Answer:

Pottasium nitrate is highly exothermic!However the dissolution of sodium nitrate in water is endothermic. Heat is absorbed during process.

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Forrest spills some acid on the lab floor. He notifies his teacher and begins to look for baking soda to neutralize the acid on
riadik2000 [5.3K]

The other students in the lab should be notified next in this type of scenario.

<h3>What is an acid?</h3>

This is a substance which donates protons and is very corrosive. It also turns blue litmus paper red.

When it was spilled and baking soda was used to neutralize it on the floor , it is best to inform the other students too so as to prevent them from being exposed by mistake thereby reducing risk of injury.

Read more about Acid here brainly.com/question/25148363

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5 0
1 year ago
Which method do you consider most appropriate for determining half-life?
Masja [62]
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3 0
3 years ago
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Harlamova29_29 [7]

Answer:

This description needs a unit describing the system of measurement.

6 0
1 year ago
A covalent bond in which electrons are shared unequally is
tekilochka [14]
Covalent bonds = sharing of electrons between two atoms of the same elements or elements close to each other on the periodic table. Usually they are metals sometimes non-metals. In polar bonds electrons are shared unequally. Non polar bonds share electrons equally.


7 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
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