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stiks02 [169]
3 years ago
12

Calculate the vapor pressure of a solution made by dissolving 109 grams of glucose (molar mass = 180.2 g/mol in 920.0 ml of wate

r at 25 °c. the vapor pressure of pure water at 25 °c is 23.76 mm hg. assume the density of the solution is 1.00 g/ml.
Chemistry
1 answer:
Fynjy0 [20]3 years ago
6 0
Given data: <span>molar mass = 180.2 g/mol in 920.0 ml of water at 25 °c.
                   </span><span>the vapor pressure of pure water at 25 °c is 23.76 mm hg. 
</span>Asked: <span>the vapor pressure of a solution made by dissolving 109 grams of glucose
</span><span>
Solution:
moles glucose = 109 g/ 180.2 g/mol=0.605 
mass water = 920 mL x 1 g/mL = 920 g 
moles water = 920 g/ 18.02 g/mol=51.1 
mole fraction water = 51.1 / 51.1 + 0.605 =0.988 
vapor pressure solution = 0.988 x 23.76 = 23.47 mm Hg</span>
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7 0
3 years ago
Convert -230,000J to kJ
MAXImum [283]

answer: -230kJ
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3 years ago
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Mila [183]

Answer: it is in the correct order

Explanation:

5 0
2 years ago
[15 Points, Stoichiometry and Gases]
KiRa [710]

The reaction is

CaC₂(s) + 2H₂O (l) -----> Ca(OH)₂ (s) + C₂H₂ (g) ​

As we have data of gas ethyne (or acetylene), C₂H₂

We can calculate the moles of acetylene and from this we can estimate the mass of calcium carbide taken

the moles of acetylene will be calculated using ideal gas equation

PV =nRT

R = gas constant = 0.0821 Latm/molK

T = 385 K

V = volume = 550 L

P = Pressure = 1.25 atm

n = moles = ?

n = PV /RT = 1.25 X 550 / 0.0821 X 385 = 21.75 mol

As per balanced equation these moles of acetylene will be obtained from same moles of calcium carbide

moles of calcium carbide = 21.75mol

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mass of CaC₂ = moles X molar mass = 21.75 X 64 = 1392g

6 0
3 years ago
Read 2 more answers
At a given temperature, the elementary reaction A − ⇀ ↽ − B , A↽−−⇀B, in the forward direction, is first order in A A with a rat
Svetllana [295]

Answer:

The equilibrium constant for the reversible reaction = 0.0164

Explanation:

At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.

The reaction is given as

A ⇌ B

Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]

The rate of forward reaction = |r₁| = k₁ [A]

The rate of backward reaction = |r₂| = k₂ [B]

(Taking only the magnitudes)

where k₁ and k₂ are the forward and backward rate constants respectively.

k₁ = 0.010 s⁻¹

k₂ = 0.0610 s⁻¹

|r₁| = 0.010 [A]

|r₂| = 0.016 [B]

At equilibrium, the rate of forward and backward reactions are equal

|r₁| = |r₂|

k₁ [A] = k₂ [B] (eqn 1)

Note that equilibrium constant, K, is given as

K = [B]/[A]

So, from eqn 1

k₁ [A] = k₂ [B]

[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164

K = [B]/[A] = (k₁/k₂) = 0.0164

Hope this Helps!!!

5 0
3 years ago
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