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a_sh-v [17]
3 years ago
5

Which of the following is the most correct statement about a gas?

Chemistry
1 answer:
matrenka [14]3 years ago
6 0
A. gasses are technically a liquid, this having relatively same qualities
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State whether the following liquids are miscible with hexane (C6H14) by entering "yes" or "no".
NeTakaya

Benzene, C6H6 is miscible with hexane (C6H14)  but iron (IIl) nitrate, and Sulfuric acid, H2SO4Fe(NO3)3 are not miscible with hexane (C6H14)

Since organic or non polar dissolves in nom polar solvent.here both benzene and hexane (C6H14) are non polar solvents.But iron (IIl) nitrate, Fe(NO3)3 and Sulfuric acid, H2SO4 are inorganic and polar, therefore cannot dissolve with non polar solvent like hexane.

Miscible is a flowery heard that oil and water aren't very miscible substances, while seltzer and orange juice are miscible and scrumptious.Miscible beverages are also described as liquids that can blend to form a homogeneous solution. Miscible drinks generally blend with out limit, meaning they may be soluble at all quantities.

Miscible approach the materials blend completely. If  substances are miscible, they're also absolutely soluble in each other no matter the order of creation. for example, tetrahydrofuran and water are miscible.

Learn more about miscible liquids here:-brainly.com/question/10423831

#SPJ1

3 0
1 year ago
How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)
Masteriza [31]

<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL    (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g

To calculate the heat absorbed by the water, we use the equation:

q=mc\Delta T

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(100-25)^oC=75^oC

Putting values in above equation, we get:

q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

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289.4 F is the right one man
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