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Arte-miy333 [17]
3 years ago
9

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

Chemistry
2 answers:
Lady_Fox [76]3 years ago
7 0

Answer:

ok

Explanation:

____ [38]3 years ago
3 0

Please dont post such questions.

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3
Mashcka [7]

Answer:

It is 34

Explanation:

4 0
3 years ago
Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th
Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M

The given equilibrium reaction is,

                           N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initially conc.      0        0           0.1576

At eqm.               (x)       (x)        (0.1576-2x)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}

By solving the term, we get:

x=0.0742,0.0839

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

5 0
3 years ago
When did the mass extinctions happen
Alisiya [41]
Out of the five major mass extinction, one is Cretaceous-Tertiary extinction (K-T extinction). It happened 66 million years ago marked by the end of the Cretinous period.
4 0
3 years ago
2/10
emmainna [20.7K]

Answer:

pH=2.34

Explanation:

HBr  -> H  + Br

The dissociation it's complete, for that reason the concentration of the products is the same of HBr

[H+]=[Br-]=0.00234 M

pH= - log (0.00234)=2.34

6 0
3 years ago
Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0
2 years ago
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