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3 years ago

Which of the following is not an antioxidant _________

Chemistry

1 answer:

Andrews [41]3 years ago

7 0

Answer: 1. Sodium Benzoate

Explanation: An anti-oxidant is a substance that can help prevent or stop the damage done by free radicals. Examples include; Sulphur Dioxide, Sulphite Salts, Citric Acid, e.t.c

Sodium benzoate is a pure preservative.

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A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

Answer: The partial pressure of oxygen gas is 2.76 bar

Explanation:

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

3 years ago

Calculate the molarity 20ml of 3.5 m kci in to a final volume of 100 ml

satela [25.4K]

Answer:

Molarity = 0.7 M

Explanation:

Given data:

Volume of KCl = 20 mL ( 0.02 L)

Molarity = 3.5 M

Final volume = 100 mL (0.1 L)

Molarity in 100 mL = ?

Solution:

Molarity = number of moles of solute / volume in litter.

First of all we will determine the number of moles of KCl available.

Number of moles = molarity × volume in litter

Number of moles = 3.5 M × 0.02 L

Number of moles = 0.07 mol

Molarity in 100 mL.

Molarity = number of moles / volume in litter

Molarity = 0.07 mol /0.1 L

Molarity = 0.7 M

3 0

3 years ago

(30 Points) You need to produce a buffer solution that has a pH of 5.02. You already have a solution that contains 10. mmol (mil

chubhunter [2.5K]

By definition:

pH = pKa + log [acetate]/ [acetic acid]

so

5.02 = 4.74 + log [acetate] / 10 mmole

10mmole = 10/1000 = 0.01 mole

5.02 = 4.74 + log [acetate] / 0.01

5.02 - 4.74 = 0.28 = log [acetate] /0.01

10^0.28 = 1.90546 = [acetate] / 0.01 
[acetate] = 0.019 mole

= 19 millimoles

8 0

3 years ago

Read 2 more answers

Question 1

nordsb [41]

#1 B

#2 C

#3 B

#4 A.

I Just did this

8 0

3 years ago