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mash [69]
2 years ago
10

Which of these would most likely be considered a pure substance?

Chemistry
2 answers:
Karolina [17]2 years ago
5 0

Answer:

c. Water Vapour

Hope it helps.

aliya0001 [1]2 years ago
4 0

the answer is C- water vapour

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Select only one most suitable description
Annette [7]
I believe it's a pure element
3 0
1 year ago
Q8. The titration of 15.00 mL of HBr solution of unknown concentration requires 18.44 mL of a 0.100 M KOH solution to reach the
lara [203]

Answer: 0.123 M

Explanation:

According to the neutralization law:

n_1M_1V_1=n_2M_2V_2

where,

M_1 = molarity of HBr solution = ?

V_1 = volume of HBr solution = 15.00 ml

M_2 = molarity of KOH solution = 0.100 M

V_2 = volume of KOH solution = 18.44 ml

n_1 = valency of HBr = 1

n_2 = valency of KOH = 1

1\times M_1\times 15.00=1\times 0.100\times 18.44

M_1=0.123

Therefore, the concentration of the unknown HBr solution is 0.123 M

8 0
2 years ago
Can alkenes react with potassium dichromate?
Lady bird [3.3K]
<span>alkenes donot react with potassium dichromate , but can react with potassium permanganate</span>
3 0
3 years ago
Complete the following conversion.<br> 25.0 °C =<br> °F
olga nikolaevna [1]

Answer: 25°C=77°F

Explanation:

FORMULA

F=9/5 C+32

-----------------------------------

Given: C=25°

F=9/5 C+32

F=9/5 (25)+32

F=45+32

F=77°

Hope this helps!! :)

Please let me know if you have any question

4 0
3 years ago
Read 2 more answers
A helium balloon with a volume of 550mL is cooled from 305 to 265K. The pressure on the gas is reduced from 0.45 atm to 0.25 atm
Aleksandr [31]

860 mL.

<h3>Explanation</h3>

Separate this process into two steps:

  1. Cool the balloon from 305 K to 265 K.
  2. Reduce the pressure on the balloon from 0.45 atm to 0.25 atm.

What would be the volume of the balloon after each step?

After Cooling the balloon at constant pressure:

By Charles's Law, the volume of a gas is directly related to its temperature in degrees Kelvins.

In other words,

\dfrac{V_2}{V_1} = \dfrac{T_2}{T_1},

where

  • V_1 and V_2 are volumes of the same gas.
  • T_1 and T_2 are the temperatures (in degrees Kelvins) of that gas.

Rearranging,

V_2 = V_1 \cdot \dfrac{T_2}{T_1}\\\phantom{V_2} = 550 \times \dfrac{265}{305}\\\phantom{V_2} = 478 \; \text{mL}.

The balloon ended up with a lower temperature. As a result, its volume drops: V_2 < V_1.

After reducing the pressure on the balloon at constant temperature:

By Boyle's Law, the volume of a gas is inversely proportional to the pressure on this gas.

In other words,

\dfrac{V_2}{V_1} = \dfrac{P_1}{P_2},

where

  • V_1 and V_2 are volumes of the same gas.
  • P_1 and P_2 are the pressures on this gas.

Rearranging,

V_2 = V_1 \cdot \dfrac{P_1}{P_2}\\\phantom{V_2} = 478 \times \dfrac{0.45}{0.25}\\\phantom{V_2} = 860 \;\text{mL}.

There's now less pressure on the balloon. As a result, the balloon will gain in volume: V_2 > V_1.

The final volume of the balloon will be 860 \; \text{mL}.

7 0
3 years ago
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