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Komok [63]
3 years ago
9

What happens to the coefficients in a reaction when writing the equilibrium constant expression?

Chemistry
1 answer:
Taya2010 [7]3 years ago
6 0

The answer is D :))..

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jok3333 [9.3K]

WHat's a concept map again?

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7 0
3 years ago
What neutral element has 17 electrons an 18 neutrons?
pishuonlain [190]

Answer:

chlorine atom

A chlorine atom has 17 protons, 18 neutrons, and 17 electrons.

Explanation:

4 0
2 years ago
Read 2 more answers
state the conditions under which copper reacts with sulphuric (vi) acid and give an equation for the reaction​
uranmaximum [27]

Answer:

When the metal reacts with hot, concentrated sulphuric acid, the products of the reaction are copper (II) sulphate, sulphur dioxide and water. Cu + 2H2SO4 = CuSO4 + SO2 + 2H2O. This is a typical redox reaction in which the acid is reduced to SO2, but no hydrogen is produced here

3 0
3 years ago
Draw a Lewis structure for C2H3Cl . Include all hydrogen atoms and show all unshared electron pairs. None of the atoms bears a f
Korvikt [17]

Answer:

See attached picture.

Explanation:

Hello!

In this case, since C2H3Cl is an organic compound we need a central C-C parent chain to which the three hydrogen atoms and one chlorine atom provides the electrons to get all the octets except for H as given on the statement.

In such a way, on the attached picture you can find the required Lewis dot structure without formal charges and with all the unshared electron pairs, considering there is a double bond binding the central carbon atoms in order to compete their octets.

Best regards!

5 0
3 years ago
At 350°c, keq = 1.67 × 10-2 for the reversible reaction 2hi (g) ⇌ h2 (g) + i2 (g). what is the concentration of hi at equilibriu
mariarad [96]
According to the reversible reaction equation:

2Hi(g) ↔ H2(g) + i2(g)

and when Keq is the concentration of the products / the concentration of the reactants.

Keq = [H2][i2]/[Hi]^2

when we have Keq = 1.67 x 10^-2

[H2] = 2.44 x 10^-3

[i2] = 7.18 x 10^-5

so, by substitution:

1.67 x 10^-2 = (2.44 x 10^-3)*(7.18x10^-5)/[Hi]^2

∴[Hi] = 0.0033 M
7 0
3 years ago
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