First calculate the mole fraction of each substance:
Acetone: 2.88 mol ÷ (2.88 mol + 1.45 mol) = 0.665
Cyclohexane: 1.45 ÷ (2.88 mol + 1.45 mol) = 0.335
Raoult's Law: P(total) = P(acetone) · χ(acetone) + P(cyclohexane) · χ(cyclohexane).
P(total) = 229.5 torr · 0.665 + 97.6 torr · 0.335
P(total) = 185.3 torr
χ for acetone: 229.5 torr · 0.665 ÷ 185.3 torr = 0.823
χ for cyclohexane: 97.6 torr · 0.335 ÷ 185.3 torr = 0.177
The pH of a 0.260 M solution of ascorbic acid is 0.585. Details about pH can be found below.
<h3>How to calculate pH?</h3>
The pH of a solution can be calculated using the following expression:
pH = - log {H+}
According to this question, ascorbic acid is a diprotic acid and posseses a concentration of 0.260M. The pH can be calculated as follows;
pH = - log {0.260}
pH = 0.585
Therefore, the pH of a 0.260 M solution of ascorbic acid is 0.585.
Learn more about pH at: brainly.com/question/15289741
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Answer: 0.174 g
Explanation:
First of all understand that charge on an atom like here 2+ will not vary molar mass. because ions are formed by loss or gain of electrons and electrons do not contribute to the mass of an atom or ion. so losing or gaining electron doesnt create any difference.
molar mass of Fe, i.e. mass of 1 mole of Fe is 56 g. So, mass of 1 mole of Fe2+ will also be 56 g.
given is 0.003109 moles of Fe2+,
Use maths,
1 mole of Fe2+ weighs 56 g
So, 0.003109 moles will weigh = (56 g / 1 mol ) x 0.003109 = 0.174 g.
To get the concentration of the second solution let us use the following formulae
C1V1=C2V2 where C1 is concentration of first solution and V1 is the volume of solution first solution. on the other hand C2 is the concentration of second solution and V2 is the volume of second solution.
therefore
0.8×2=(2+10)×C2
1.6 =12×C2
1.6/12=C2
C2 = 0.1333mg/mL
Answer:
0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidised. ... Thus, 90g of lower oxide contains as much metal as 100g of higher oxide, i.e., 80g (given). Hence, 80g of metal combines with 10g of oxygen in the lower oxide and 20g of oxygen in the higher oxide.
