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prohojiy [21]
3 years ago
6

What is the osmotic pressure of a solution made by dissolving 75.0 g of glucose, c6h12o6, in enough water to form 700.0 ml of so

lution at 45.0 ∘c ? express your answer to three significant figures and include the appropriate u?
Chemistry
1 answer:
lions [1.4K]3 years ago
8 0
We will use the osmotic pressure formula:
π = n R T / V
When π is the osmotic pressure (atm)
n is no.of moles when the molar mass of glucose = 180 g/mol
so, n = 75 g / 180 g/mol= 0.42

and R is gas constant = 0.0821 L.atm/mol.k 
T is the temperature in Kelvin = 45 +273 = 318 K
and V is the volume in Litre = 700 / 1000 = 0.7 L
So, by substitution:
∴ π = 0.42 * 0.0821 * 318 / 0.7 = 15.665 atm
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39.1 gCO2
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3 years ago
What is the reaction quotient, Q, for this system when [N2] = 2.00 M, [H2] = 2.00 M, and [NH3] = 1.00 M at 472°C?
cupoosta [38]

Answer : The value of reaction quotient, Q is 0.0625.

Solution : Given,

Concentration of N_2 = 2.00 M

Concentration of H_2 = 2.00 M

Concentration of NH_3 = 1.00 M

Reaction quotient : It is defined as a concentration of a chemical species involved in the chemical reaction.

The balanced equilibrium reaction is,

N_2+3H_2\rightleftharpoons 2NH_3

The expression of reaction quotient for this reaction is,

Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}

Now put all the given values in this expression, we get

Q=\frac{(1.00)^2}{(2.00)^1(2.00)^3}=0.0625

Therefore, the value of reaction quotient, Q is 0.0625.

3 0
3 years ago
HELP! ASAP! Iron (Fe) and copper (II) chloride (CuCl2) combine to form iron (III) chloride (FeCl3) and copper (Cu). If you start
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Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each
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Answer:

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Reduced: O₂

Oxidized : C₃H₈

Explanation:

For the given reaction:

C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(l)

In the reactants, there are 3 C, and in the products only one, so we multiply CO₂ by 3:

C₃H₈(g) + O₂(g) → 3CO₂(g) + H₂O(l)

In the reactants, there are 8 H, and in the products, there are 2 H, so we multiply H₂O by 4:

C₃H₈(g) + O₂(g) → 3CO₂(g) + 4H₂O(l)

In the reactants are 2 O, and in the products 10 O, so we multiply O₂ by 5:

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The Nox of O is fix equal to -2, and the molecule is neutral, so, calling x the Nox of C:

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H₂O:

The Nox of H and O are fixed, respectively, +1 and -2.

So, the carbon in C₃H₈ is oxidized because its Nox is increasing, and oxygen in O₂ is reduced because its Nox is decreasing.

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