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kow [346]
3 years ago
8

______ is the type of heat that roasts the marshmallows.

Chemistry
1 answer:
disa [49]3 years ago
5 0

Answer:

I think B

tell me if I'm wrong

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Examples of polar covalent bonds
mars1129 [50]
A water molecule, abbreviated as H2O, is an example of a polar covalent bond. The electrons are unequally shared, with the oxygen atom spending more time with electrons than the hydrogen atoms. Since electrons spend more time with the oxygen atom, it carries a partial negative charge.
8 0
2 years ago
What is the molarity of H3PO4?
masya89 [10]
The answer is 14.8 hoped this helped
3 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
To name the compound written as CuCl2, you would write:
vovangra [49]

Answer: option C. Copper (II) chloride

Explanation:

To name CuCl2, we need to know the oxidation state of Cu in the compound as chlorine always have oxidation on —1 in all its compound. The oxidation state of Cu can be calculated as follows:

Cu + 2Cl = 0 (since the compound has no charge)

Cl = —1

Cu + 2(—1) = 0

Cu —2 = 0

Collect like terms

Cu = 0 +2

Cu = +2

Therefore, the oxidation state of Cu in CuCl2 is +2.

The name of the compound will be copper(ii) chloride, since cupper has oxidation state +2 in the compound.

4 0
2 years ago
C12H7Cl3FNaO2 what are them elements in that chemical formula
LuckyWell [14K]
Assuming you are asking for the names of the elements in that formula , the answer is

carbon
hydrogen
chlorine
fluorine
sodium
oxygen
8 0
3 years ago
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