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3 years ago

Look up and record the boiling point of acetic acid, and explain why only some of it evaporates from the reaction mixture.

Chemistry

1 answer:

REY [17]3 years ago

8 0

Answer:

Heating the mixture to a temperature above the boiling point of acetic acid, but below 100°C (the boiling point of water). The vapours from the acetic acid rise, and go into a tube. They are then condensed within the tube, and run off into a separate storage area. Because water can exist as a gas at pretty much any temperature above 0°C, it will result in an impure mixture, but repeatedly doing this will get the acetic acid to the desired purity.

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8.632 nm + 8.3 nm - 30.0 nm=

Misha Larkins [42]

Answer:

13.06800 nanometers

Explanation:

i hope you had understand !

5 0

3 years ago

Please help me would really appreciate it thank you :)

tatyana61 [14]

Answer:

e. 8.04*10^4

Explanation:

80.4 g converted to mg is 80,400. 8.04*10^4= 80,400

4 0

2 years ago

Which squirrel do you think is better suited for the environment, the gray one or the white one? and why

boyakko [2]

Answer:

The grey squirrel.

Explanation:

This is your answer because the white one would have a harder time to blend in with the tree's and the environment, whilst the grey one is better suited for its environment better due to the fact that it can blend in with the trees, and hide better.

3 0

2 years ago

How many kilograms of potassium iodide (ki) are needed to make 1.25 l of a 4.41 m ki solution?

Lapatulllka [165]

C = n/V
n = C×V
n = 4,41M × 1,25L
n = 5,5125 mol

mKI: 39+127 = 166 g/mol

1 mol --------- 166g
5,5125 mol --- X
X = 166×5,5125 = 915,075g KI

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7 0

3 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

5 0

2 years ago