Which of the following is equal to 0.000539?

List three ways the rate of solvation of sodium chloride in water may be<br> increased

r-ruslan [8.4K]

Answer:

1) Increasing temperature

2) Stirring

3) Increasing surface area of salt by grinding it

5 0

2 years ago

Suppose that Daniel has a 3.00 3.00 L bottle that contains a mixture of O 2 O2 , N 2 N2 , and CO 2 CO2 under a total pressure of

Alenkinab [10]

Answer:

Partial pressure O₂ → 2.74 atm

Explanation:

Let's analyse the data given:

Volume → 3L

In the bottle there is a mixture of gases that contains, O₂, N₂ and CO₂.

Total pressure is 4.80 atm

Let's apply the Ideal Gases Law to determine the total moles of the mixture

P . V = n . R. T

4.80 atm . 3L = n . 0.082 . 273K

n = 4.80 atm . 3L / 0.082 . 273K → 0.643 moles

We apply the concept of mole fraction:

Mole fraction of a gas X = moles of gas X / Total moles

Mole fraction of a gas X = Partial pressure X / Total pressure

In a mixture, sum of mole fraction of each gas = 1

We determine mole fraction of N₂ → 0.230 / 0.643 = 0.357

We determine mole fraction of CO₂ → 0.350 atm / 4.80 atm = 0.0729

1 - mole fraction N₂ - mole fraction CO₂ = mole fraction O₂

1 - 0.357 - 0.0729 = 0.5701 → mole fraction O₂

We replace in the formula: Mole fraction O₂ = Partial pressure O₂ / 4.80 atm

0.5701 . 4.80 atm = Partial pressure O₂ → 2.74 atm

5 0

2 years ago

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The complementary base sequence to GTAGCT is:_______.A) CATCGA.B) ACGATC.C) GCAGTA.D) AGTCGA.E) None of the above

IceJOKER [234]

Answer:none of the above

Explanation:

8 0

2 years ago

Please I need help on these 3 questions. Thank You.

Liula [17]

1.

V = 200 mL (volume)

c = 3 M = 3 mol/L (concentration)

First we convert mL to L:

200 mL = 0.2 L

Then we calculate the moles using the formula: n = V × c = 0.2 L × 3 mol = 0.6 mol

Finally, we just use the molar mass of CaF2 to calculate the actual mass:

molar mass = 78 g/mol

The formula is: m = n × mm (mass = moles × molar mass)

m = 0.6 mol × 78 g/mol = 46.8 g

2.

For this question the steps are exactly like the first question.

V = 50mL = 0.05 L

c = 12 M = 12 mol/L

n = V × c = 0.05 L × 12 mol/L = 0.6 mol

molar mass (HCl) = 36.5 g/mol

m = n × mm = 0.6 mol × 36.5 g/mol = 21.9 g.

3.

The steps for this question are the opposite way.

m(K2CO3) = 250 g

molar mass = 138 g/mol

n = m ÷ mm = 1.81 mol

c = 2 mol/L

V = n ÷ c = 1.81 mol ÷ 2 mol/L = 0.905 L = 905 mL

6 0

3 years ago

What is the pH of 1.0M HI solution?

ozzi

Answer:

0

Explanation:

Since HI is a strong acid, the amoung of Hydrogen ions produced by it will be the same molar as the reactant. The negative log of the concentration will reveal that the pH is 0.

7 0

3 years ago

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