Is aloe vera a living or non-living thing?

When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

3 0

3 years ago

Which of the following solids would not decompose on heating

Gekata [30.6K]

Answer:

Anhydrous sodium carbonate is stable to heat and does not decompose even when it is heated to redness. This is because sodium carbonate salt on heating with acids react to release carbon dioxide.

5 0

3 years ago

Consider the following intermediate reactions.

Alja [10]

2.1648 kg of CH4 will generate 119341 KJ of energy.

Explanation:

Write down the values given in the question

CH4(g) +2 O2 → CO2(g) +2 H20 (g)

ΔH1 = - 802 kJ

2 H2O(g)→2 H2O(I)

ΔH2= -88 kJ

The overall chemical reaction is

CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ

CH4 +2 O2 → CO2 +2 H20

(1mol)+(2mol)→(1mol+2mol)

Methane (CH4) = 16 gm/mol

oxygen (O2) =32 gm/mol

Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O

which generate 882 KJ /mol

Therefore to produce 119341 KJ of energy

119341/882 = 135.3 mol

to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require

=135.3 *16

=2164.8 gm

=2.1648 kg of CH4

2.1648 kg of CH4 will generate 119341 KJ of energy

4 0

3 years ago

Read 2 more answers

Explain what is meant by energy and list two units in which energy is measured

dybincka [34]

Answer:

J (joule) W(watt)

Explanation:

If you're looking for the definition it is

The most common definition of energy is the work that a certain force can do. Energy also cannot be created or destroyed and some examples are

light, heat, mechanical, potential, and kinetic

4 0

3 years ago

7. Suppose 1.01 g of iron (III) chloride is placed in a 10.00-mL volumetric flask with a bit of water in it. The flask is shaken

Nana76 [90]

<u>Answer:</u> The molarity of Iron (III) chloride is 0.622 M.

<u>Explanation:</u>

Molarity is defined as the number of moles present in one liter of solution. The equation used to calculate molarity of the solution is:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Or,

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of iron (III) chloride = 1.01 g

Molar mass of iron (III) chloride = 162.2 g/mol

Volume of the solution = 10 mL

Putting values in above equation, we get:

$\text{Molarity of Iron (III) chloride}=\frac{1.01g\times 1000}{162.2g/mol\times 10mL}\\\\\text{Molarity of Iron (III) chloride}=0.622M$

Hence, the molarity of Iron (III) chloride is 0.622 M.

3 0

3 years ago