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Chemistry

1 answer:

yulyashka [42]1 year ago

6 0

Answer:

I hope this helps you :)

Explanation:

Question 1 - B
Question 2 - A
Question 3 - D
Question 4 - C

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Four most common oxyanions of Iodine

nikitadnepr [17]

HIO= Hypoiodous acid; IO-=Hypoiodite
HIO2= Iodous acid; IO2= iodite
HIO3=Iodic acid; IO3-= iodate
HIO4=Periodic acid; IO4-=Periodate

8 0

2 years ago

Read 2 more answers

If 40.0 g of HCl react with an excess of magnesium metal, what is the theoretical yield of hydrogen?

lyudmila [28]

The balanced reaction is:<span>

Mg (s) + 2HCl (aq) --> MgCl 2 (aq) + H 2 (g)

We are given the amount hydrochloric acid to be used for the reaction. This will be the starting point of the calculation.

40.0 g HCl ( 1 mol HCl / 36.46 g HCl) (1 mol H2 / 2 mol HCl) (2.02 g H2 / 1 mol H2) = 1.11 g H2</span>

8 0

2 years ago

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.600 M. N2(g)+O2(g)↽−−⇀2NO(g) If

cestrela7 [59]

Answer:

0.84M

Explanation:

Hello,

At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:

$K_{eq}=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}} \\K_{eq}=\frac{0.6^2}{0.2*0.2}\\ K_{eq}=9$

Now, the new equilibrium condition, taking into account the change x, becomes:

$9=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}}\\9=\frac{[0.9+2x]^2}{[0.2-x][0.2-x]}$

Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:

$\frac{1}{9} =\frac{[N_2]_{eq}[O_2]_{eq}}{[NO]^2_{eq}}\\\frac{1}{9} =\frac{[0.2+x][0.2+x]}{[0.9-2x]^2}$