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Anettt [7]
4 years ago
11

The molar enthalpy of

Chemistry
1 answer:
nika2105 [10]4 years ago
7 0

Answer:

vaporization

Explanation:

The molar enthapy of _vaporization______ is the heat required to vaporize one mole of a liquid”

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Please help me answer this ASAP
hoa [83]

Answer:

h B i f M r

Explanation:

that was just my estimated guess but it seems right to me

4 0
3 years ago
Read 2 more answers
Determine the composition of each phase in a cu-40% ni alloy at 1300c,
Morgarella [4.7K]
To be able to answer this, you need to have a phase diagram of Cu-Ni alloy. Create a vertical line along 40% in the x-axis. Then, create a horizontal line along the temperature at 1,300°C. The intersection lies outside of the α-L boundary. <em>Therefore, the system consist solely of the L-phase: 100% L.</em>

4 0
3 years ago
A buffer solution is prepared by adding 13.74 g of sodium acetate (NaC2H3O2) and 15.36 g of acetic acid to enough water to make
hoa [83]

Answer:

A buffer solution is prepared by adding 13.74 g of sodium acetate (NaC2H3O2) and 15.36 g of acetic acid to enough water to make 500 mL of solution.

Calculate the pH of this buffer.

Explanation:

The pH of a buffer solution can be calculated by using the Henderson-Hesselbalch equation:

pH=pKa+log\frac{[salt]}{[acid]}

The pH of the given buffer solution can be calculated as shown below:

6 0
4 years ago
Your lab partner combined chloroform and acetone to create a solution where the mole fraction of chloroform, Xchloroform, is 0.1
jeyben [28]

Answer:

Explanation:

[u]Assumptions[/u]

1. There is exactly 1 mole of chloroform

2. The liquids mix together well such that the volume of the solution is the sum of the volumes of the two liquids

Given that the mole fraction of the Chloroform is 0.171

Mole fraction of Chloroform =

Mole of chloroform /(Mole of chloroform + mole of acetone)

According to assumptions, mole of chloroform is equal to 1

Therefore 0.171 =1/(1+mole of acetone)

1 + mole of acetone = 1/0.171

Mole of acetone = 1(/0.171) - 1

Moles of acetone = 4.85mol.

From Stochiometry

Mass of acetone = Mole of acetone * Molar Mass of acetone

Molar mass of acetone = 58.1grams/mol

Mass of acetone = 4.85 *58.1 = 282g =0.282kg

Mass of chloroform = moles of chloroform *Molar mass of Chloroform

Molar mass of chloform = 119.4 grams/mol

Mass of chloroform = 1* 119.4 =119.4g=0.1994kg

Volume of acetone = Mass of acetone / Density of acetone

Volume of acetone = 282/0.791

Volume of acetone = 357mL

Volume of Chloroform = Mass of Chloroform /Density of Chloroform

Volume of Chloroform = 119.4/1.48

= 81mL

Total volume of solution = 357mL+81mL = 438mL = 0.438L

1. Molarity = Moles of solute(chloroform)/mass of solvent (acetone) in kg

Molarity = 1/0.282 = 3.55molal

2. Molarity = moles of solute( chloroform) /Volume of solution

= 1/0.438 =2.28Molar

Therefore the molality and molarity respectively are 3.55 and 2.28.

4 0
4 years ago
When silver nitrate is added to an aqueous solution of magnesium chloride, a precipitation reaction occurs that produces silver
Aloiza [94]

Answer:

103.62 g of AgCl.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AgNO3 + MgCl2 —> 2AgCl + Mg(NO3)2

Step 2:

Determination of the mass of MgCl2 that reacted and the mass of AgCl produced from the balanced equation.

This is illustrated below:

Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol

Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g

Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Mass of AgCl from the balanced equation = 2 x 143.5 = 287 g

Thus, from the balanced equation above,

95 g of MgCl2 reacted to produce 287 g of AgCl.

Step 3:

Determination of the mass of AgCl produced from the reaction of 34.3 g of MgCl2.

The mass of AgCl produced from the reaction can be obtained as follow:

Form the balanced equation above,

95 g of MgCl2 reacted to produce 287 g of AgCl.

Therefore, 34.3 g of MgCl2 will react to produce = (34.3 x 287)/95 = 103.62 g of AgCl.

Therefore, 103.62 g of AgCl were produced from the reaction.

8 0
4 years ago
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