What would be the number of oxygen atoms in 5.00 moles of co2?

A manganese electrode was oxidized electrically. If the mass of the electrode decreased by 225 mg during the passage of 1580 cou

qwelly [4]

Answer:

Oxidation state of Mn = +4

Explanation:

Atomic mass of Mn = 55g/mol

From Faraday's law of electrolysis,

Electrochemical equivalent = $\frac{mass}{charge}$

i.e Z = $\frac{m}{Q}$ = $\frac{0.225g}{1580C}$ = 0.0001424 g/C

But Equivalent weight, E = atomic mass ÷ valency = Z × 96,485

⇒ $\frac{55}{valency}$ = 0.0001424 × 96,485

∴ Valency of Mn = +4

8 0

3 years ago

Pls help need it now as I don't know the answer

ioda

Answer:

Iron :)))))))))))))))))))))))))

3 0

3 years ago

Read 2 more answers

Two plates slide past each other in opposite directions at a ____ boundary.

Kitty [74]

Answer: plate

Explanation: Plates grinding past each other in opposite directions create faults called transform faults. Powerful earthquakes often strike along these boundaries. The San Andreas Fault is a transform plate boundary that separates the North American Plate from the Pacific Plate.

4 0

3 years ago

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Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol

pishuonlain [190]

Answer: The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

Explanation:

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

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

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

$0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g$

Calculating the freezing point:

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

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = -114.1 °C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.99°C/m

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC$

Hence, the freezing point of solution is -117.54°C

Calculating the boiling point:

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:

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

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

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

Boiling point of pure solution = 78.4°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_b$ = molal boiling point elevation constant = 1.20°C/m.g

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-78.4=1\times 1.20^oC/m\times \frac{8.15\times 1000}{62\times 75.98}\\\\\text{Boiling point of solution}=80.48^oC$

Hence, the boiling point of solution is 80.48°C

3 0

4 years ago

(a) The student dissolves the entire impure sample of CuSO4(s) in enough distilled water to make 100.mL of solution. Then the st

Talja [164]

Answer:

Explanation:

Given parameters :

Volume of solution = 100mL

Absorbance of solution = 0.30

Unknown:

Concentration of CuSO₄ in the solution = ?

Solution:

There is relationship between the absorbance and concentration of a solution. They are directly proportional to one another.

A graph of absorbance against concentration gives a value of 0.15M at an absorbance of 0.30.

The concentration is 0.15M

Also, we can use: Beer-Lambert's law;

A = ε mC l

where εm is the molar extinction coefficient

C is the concentration

l is the path length

Since the εm is not given and assuming path length is 1;

Then we solve for the concentration.

3 0

3 years ago