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vesna_86 [32]
2 years ago
7

When 2.5 mol of O2 are consumed in their reaction, ________ mol of CO2 are produced

Chemistry
1 answer:
Vika [28.1K]2 years ago
4 0

The given question is incomplete. The complete question is:

The combustion of propane (C3H8) in the presence of excess oxygen yields CO_2 and H_2O

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

When only 2.5 mol of O_2 are consumed in order to complete the reaction, ________ mol of CO_2 are produced.

Answer: Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

Explanation:

The balanced chemical equation is:

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

According to stoichiometry :

5 moles of O_2 produce = 3 moles of  CO_2

Thus 2.5 moles of O_2 will produce = \frac{3}{5}\times 2.5=1.5 moles of  CO_2

Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

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determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass
Blababa [14]

The freezing point depression of a solution containing 30.7 g of glycerin  is  calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

K_{f}or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

             ΔT_{f} = i ×K_{f} ×m

T⁰-T_{s}  = i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water     in     kg}

in which,

ΔT_{f} = change in freezing point

ΔT_{s} = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T_{s} = 1 ×1.86°C/m ×\frac{30.7}{92.09g/mol * 0.376kg}

T_{s} = -1.65°C

Thus, the freezing point depression of a solution is  -1.65°C

<h2 />

Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

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8 0
10 months ago
How many sigma bonds does CS2 have?
Pie

Answer:

2 sigma bonds

CS2 is called carbon disulphide. The 2 unpaired electrons of each sulphur atom get paired with the 4 unpaired electrons of carbon. But it is a double bond. That is, 2 sigma bonds (sigma s-p , sigma p-p) and 2 pi (πp-p) bonds.

Explanation:

8 0
3 years ago
A gas is collected in a 34.3L container at a temperature of 31.5°C. Later, the container has a volume of 29.2L, a temperature of
goldenfox [79]

Answer:

108 kPa  

Step-by-step explanation:

To solve this problem, we can use the <em>Combined Gas Laws</em>:

p₁V₁/T₁ = p₂V₂/T₂             Multiply each side by T₁

   p₁V₁ = p₂V₂ × T₁/T₂      Divide each side by V₁

      p₁ = p₂ × V₂/V₁ × T₁/T₂

Data:

p₁ = ?;                 V₁ = 34.3 L; T₁ = 31.5 °C

p₂ = 122.2 kPa; V₂ = 29.2 L; T₂ = 21.0 °C

Calculations:

(a) Convert temperatures to <em>kelvins </em>

T₁ = (31.5 + 273.15) K = 304.65 K

T₂ = (21.0 + 273.15) K = 294.15 K

(b) Calculate the <em>pressure </em>

p₁ = 122.2 kPa × (29.2/34.3) × (304.65/294.15)  

   = 122.2 kPa × 0.8542 × 1.0357

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