How many neutrons does an element have if its atomic number is 56 and its mass number is 152?

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

If 0.25 moles of KBr is dissolved in 0.5 liters of

Len [333]

Answer:

[KBr] = 454.5 m

Explanation:

m is a sort of concentration that indicates the moles of solute which are contianed in 1kg of solvent.

In this case, the moles of solute are 0.25 moles.

Let's determine the mass of solvent in kg.

Density of heavy water, solvent, is 1.1 g/L and our volume is 0.5L.

1.1 g = mass of solvent / 0.5L, according to density.

mass of solvent = 0.5L . 1.1g/L = 0.55 g

We convert the mass to kg → 0.55 g . 1kg /1000g = 5.5×10⁻⁴ kg

m = mol/kg → 0.25 mol /5.5×10⁻⁴ kg = 454.5 m

6 0

2 years ago

For the following reaction, 9.30 grams of glucose (C6H12O6) are allowed to react with 13.8 grams of oxygen gas. glucose (C6H12O6

amid [387]

Answer:

13.7 g of CO₂

Limiting reactant: C₆H₁₂O₆

3.81 g of O₂

Explanation:

We convert the mass of the reactants to moles, in order to find out the limiting reactant and the excess reagent

9.30 g / 180 g/mol = 0.052 moles of glucose

13.8 g / 32 g/mol = 0.431 moles of oxygen

The equation is: C₆H₁₂O₆(s) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l)

Ratio is 1:6. Let's consider this rule of three:

1 mol of glucose reacts with 6 moles of oxygen

Then, 0.052 moles of glucose must react with (0.052 . 6) /1 = 0.312 moles

We have 0.431 moles of oxygen and we only need 0.312 moles. This means that an amount of oxygen still remains after the reaction is complete:

0.431 - 0.312 = 0.119 moles. We convert the moles to mass:

0.119 mol . 32 g / 1mol = 3.81 g

In conclussion, the limiting reactant is the glucose.

6 moles of oxygen react with 1 mol of glucose

0.431 moles of O₂ will react with (0.431 . 1) /6 = 0.072 moles of glucose

We only have 0.052 moles, so it is ok to say, that glucose is the limiting cause we do not have enough glucose.

Let's verify, the maximum amount of carbon dioxide that can be formed:

1 mol of glucose can produce 6 moles of CO₂

Therefore 0.052 moles of glucose will produce (0.052 . 6) /1 = 0.312 moles

We convert the moles to mass → 0.312 mol . 44 g /1 mol = 13.7 g

6 0

3 years ago

5. Hydrogen & oxygen react chemically to form water. How much water

stepladder [879]

39.25 g of water (H₂O)

Explanation:

We have the following chemical reaction:

2 H₂ + O₂ → 2 H₂O

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of H₂ = 14.8 / 2 = 7.4 moles

number of moles of O₂ = 34.8 / 32 = 1.09 moles

We see from the chemical reaction that 2 moles of H₂ will react with 1 mole of O₂ so 7.4 moles of H₂ will react with 3.7 moles of O₂ but we only have 1.09 moles of O₂ available. The O₂ will be the limiting reactant. Knowing this we devise the following reasoning:

if 1 moles of O₂ produces 2 moles of H₂O

then 1.09 moles of O₂ produces X moles of H₂O

X = (1.09 × 2) / 1 = 2.18 moles of H₂O

mass = number of moles × molar weight

mass of H₂O = 2.18 × 18 = 39.25 g

Learn more about:

limiting reactant

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6 0

2 years ago