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3 years ago

If 7.83 g of H 2 is reacted with 15.7 g of O 2 in the reaction 2H2 + O2 → 2H2O , what is the limiting reagent?

1 answer:

8 0

When you take 7.83 g of H2, you convert to moles by dividing by the molar mass (2.02) and multiply by the number of H2s over H2Os. Then do the same for the O2. the limiting reagent in this case is the Oxygen by what I calculated.

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A solution was prepared by dissolving 2.2 g of an unknown solute in 16.7 g of CCl4. A thermal analysis was performed for this so

laila [671]

Answer:

Molar mass of unknown solute is 679 g/mol

Explanation:

Let us assume that the solute is a non-electrolyte.

For a solution with non-electrolyte solute remains dissolved in it -

Depression in freezing point of solution, $\Delta T_{f}=K_{f}.m$

where, m is molality of solute in solution and $K_{f}$ is cryogenoscopic constant of solvent.

Here $\Delta T_{f}=(-22.9^{0}\textrm{C})-(-28.7^{0}\textrm{C})=5.8^{0}\textrm{C}$

If molar mass of unknown solute is M g/mol then-

$m=\frac{\frac{2.2}{M}}{0.0167}mol/kg$

So, $5.8^{0}\textrm{C}=29.9^{0}\textrm{C}/(mol/kg)\times \frac{\frac{2.2g}{M}}{0.0167}mol/kg$

so, M = 679 g/mol

4 0

3 years ago

ASAP

sammy [17]

The answer is C.

The vast difference in electronegativity of the oxygen and hydrogen in water, the O-H bond is polar.

8 0

2 years ago

Using the model of the periodic table, which two elements pictured have similar chemical properties?

Murljashka [212]

Answer:

1 and 3

Explanation:

The vertical columns (groups) of the periodic table are arranged such that all its elements have the same number of valence electrons. All elements within a certain group thus share similar properties.

5 0

3 years ago

Read 2 more answers

How many grams of argon would it take to fill a large light bulb with a volume of 0.745 L at STP?

EastWind [94]

Answer:

Mass = 1.33 g

Explanation:

Given data:

Mass of argon required = ?

Volume of bulb = 0.745 L

Temperature and pressure = standard

Solution:

We will calculate the number of moles of argon first.

Formula:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

By putting values,

1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K

0.745 atm. L = n × 22.43 atm.L/mol

n = 0.745 atm. L / 22.43 atm.L/mol

n = 0.0332 mol

Mass of argon:

Mass = number of moles × molar mass

Mass = 0.0332 mol × 39.95 g/mol

Mass = 1.33 g

8 0

3 years ago

A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and

lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

$F_{solution}$ = mass of solvent + mass of oil (i.e A+C)

Feed Phase:

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

= 25kg

Mass ratio of oil to solution $Y_{F}$ = $\frac{Mass C}{Mass (A+C)}$

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg

therefore $Y_{F}$ = $\frac{2.5}{25}$

= 0.1

mass ratio of solid to solution $Y_{A}$ = $\frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}$

Solvent Phase:

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

Underflow:

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = $\frac{mass B}{mass(A+C)}$

solution in underflow E₁ = Mass (A+C)

Overflow:

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75= E₁ × N₁

75 = E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

5 0

3 years ago