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gtnhenbr [62]
3 years ago
10

The vapor pressure of a system at equilibrium

Chemistry
1 answer:
Marizza181 [45]3 years ago
3 0

Answer: The vapor pressure or equilibrium vapor pressure is defined as the pressure exerted by a vapor that is in thermodynamic equilibrium with the condensed phase (solid or liquid) at a given temperature in a closed system. The equilibrium vapor pressure is an indication of the evaporation rate of a liquid.

Explanation:

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How is fractional distillation of crude oil related to stoichiometry
lora16 [44]

Answer:

Fractional distillation is the process by which oil refineries separate crude oil into different, more useful hydrocarbon products based on their relative molecular weights in a distillation tower.

8 0
3 years ago
An aqueous solution of sodium hydroxide is standardized by titration with a 0.170 M solution of perchloric acid. If 28.5 mL of b
Ray Of Light [21]

Answer:

0.095M

Explanation:

HClO4 + NaOH = NaClO4 + H2O

Concentration of acid CA= 0.170M

Concentration of base CB= ???

Volume of base VB= 28.5ml

Volume of acid VA= 16.0ml

Number of moles of acid nA= 1

Number of moles of base nB= 1

From

CAVA/CBVB= nA/nB

CB= CAVAnB/VBnA

CB= 0.170×16.0×1/28.5×1

CB= 0.095M

4 0
3 years ago
Can someone please help me with this, i'll give you a 5 star rating and the brainliest answer!​
elena-s [515]

Answer:

A. equation

B. atoms

C. products

Explanation:

4 0
3 years ago
Read 2 more answers
Reading the temperature of a solution by using a thermometer is an example of a(n) ________.
weeeeeb [17]

Answer:

i guess its example of observation

6 0
3 years ago
Consider the reaction ch3oh(l)→ch4(g)+1/2o2(g). part a calculate δrh∘298.
Vsevolod [243]
To determine the standard heat of reaction, ΔHrxn°, let's apply the Hess' Law.

ΔHrxn° = ∑(ν×ΔHf° of products) - ∑(ν×ΔHf° of reactants)
where
ν si the stoichiometric coefficient of the substances in the reaction
ΔHf° is the standard heat of formation

The ΔHf° for the substances are the following:
CH₃OH(l) = -238.4 kJ/mol
CH₄(g) = -74.7 kJ/mol
O₂(g) = 0 kJ/mol

ΔHrxn° = (1 mol×-74.7 kJ/mol) - ∑(1 mol×-238.4 kJ/mol)
ΔHrxn° = +163.7 kJ
8 0
3 years ago
Read 2 more answers
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