Products of MnO + CL2

When 8.70 kJ of thermal energy is added to 2.50 mol of liquid methanol, it vaporizes. Determine the heat of vaporization in kJ/m

Anni [7]

Answer:

The heat of vaporisation of methanol is "3.48 KJ/Mol"

Explanation:

The amount of heat energy required to convert or transform 1 gram of liquid to vapour is called heat of vaporisation

When 8.7 KJ of heat energy is required to vaporize 2.5 mol of liquid methanol.

Hence, for 1 mol of liquid methanol, amount of heat energy required to evaporate the methanol is = $\frac{8.7}{2.5}KJ$

= 3.48 KJ

So, the heat of vaporization $\delta H_{vap} = 3.48KJ/Mol$

Therefore, the heat of vaporization of methanol is 3.48KJ/Mol

5 0

3 years ago

What is the empirical formula of a compound consisting of 29.6% oxygen and 70.4% fluorine by mass?

Rainbow [258]

O1Fl2

1. Assume an 100g sample, so the percentage will stay the same

2. Covert each element into their molar mass
29.6/16.00=1.8 mols of O
70.4/19.00=3.7 mols of Fl

3. Divide both by the smallest value of mol
1.8/1.8=1 O
3.7/1.8=2 Fl

4. Write the empirical formula:
O1Fl2

3 0

3 years ago

Read 2 more answers

Why don't noble gasses want to react with anyone?

Zepler [3.9K]

They have a full valence shell (8 electrons in the outer shell) so they are the most stable elements on the periodic table. Therefore, they do not need to react to other elements to gain or lose elections to become stable.

5 0

3 years ago

Formula of a copper (II)sulfate hydrate lab

s344n2d4d5 [400]

Answer:

Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulphate. Record all weighings accurate to the nearest 0.01 g.

Support the crucible securely in the pipe-clay triangle on the tripod over the Bunsen burner.

Heat the crucible and contents, gently at first, over a medium Bunsen flame, so that the water of crystallisation is driven off steadily. The blue colour of the hydrated compound should gradually fade to the greyish-white of anhydrous copper(II) sulfate. Avoid over-heating, which may cause further decomposition, and stop heating immediately if the colour starts to blacken. If over-heated, toxic or corrosive fumes may be evolved. A total heating time of about 10 minutes should be enough.

Allow the crucible and contents to cool. The tongs may be used to move the hot crucible from the hot pipe-clay triangle onto the heat resistant mat where it should cool more rapidly.

Re-weigh the crucible and contents once cold.

Calculation:

Calculate the molar masses of H2O and CuSO4 (Relative atomic masses: H=1, O=16, S=32, Cu=64)

Calculate the mass of water driven off, and the mass of anhydrous copper(II) sulfate formed in your experiment

Calculate the number of moles of anhydrous copper(II) sulfate formed

Calculate the number of moles of water driven off

Calculate how many moles of water would have been driven off if 1 mole of anhydrous copper(II) sulfate had been formed

Write down the formula for hydrated copper(II) sulfate.

#*#*SHOW FULLSCREEN*#*#

Explanation:

7 0

2 years ago

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago