How did the gold foil experiment lead to the discovery of the nucleus?

What is the pOH of a solution with [OH-] = 1.4 x 10-13?

lukranit [14]

Answer:

C. 12.85

Explanation:

just did

4 0

2 years ago

Elements consist of tiny particles called

Nutka1998 [239]

The answer is atoms good luck

4 0

2 years ago

Calculate the pOH of a 0.0143 M NaOH solution at 25 ⁰C.

arsen [322]

Answer:

A.) 1.845

Explanation:

You can find the pOH using the following equation:

pOH = -log[OH⁻]

Since NaOH dissociates into 1 Na⁺ and 1 OH⁻, the concentration of both ions is 0.0143 M.

pOH = -log[OH⁻]

pOH = -log[0.0143]

pOH = 1.845

5 0

2 years ago

Salt is often added to water to raise the boiling point to heat food more quickly. if you add 30.0g of salt to 3.75kg of water,

sammy [17]

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

<h3>What is the boiling-point elevation?</h3>

Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.

Step 1: Calculate the molality of the solution.

We will use the definition of molality.

b = mass solute / molar mass solute × kg solvent

b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m

Step 2: Calculate the boiling-point elevation.

We will use the following expression.

ΔT = Kb × m × i

ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C

where

ΔT is the boiling-point elevation
Kb is the ebullioscopic constant.
b is the molality.
i is the Van't Hoff factor (i = 2 for NaCl).

The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:

100 °C + 0.140 °C = 100.14 °C

Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.

Learn more about boiling-point elevation here: brainly.com/question/4206205

7 0

2 years ago

A 0.964 gram sample of a mixture of sodium formate and sodium chloride is analyzed by adding sulfuric acid. The equation for the

frosja888 [35]

Answer: 67.8 %.

Explanation:

Okay, let us delve right into the solution to the question;

The balanced chemical reaction is given by the equation (1) below;

2 HCOONa + H2SO4 ---------> 2 CO + 2 H2O + Na2SO4. ----------------------------------------------------------------------------(1).

From the balanced chemical reaction in equation (1) above we can see that; 2 moles of HCOONa reacts with one moles of tetraoxosulphate acid, H2SO4 to produce 2 moles of carbonmonoxide,CO; 2 moles of water, H2O and 1 mole of sodium tetraoxosulphate, Na2SO4.

The parameters given from the question are; total atmospheric pressure, P(t) = 752 torr, volume of CO= 242 mL = 0.242 Litres.

STEP ONE : find the carbon monoxide,CO pressure; P(CO).

Using the formula below;

P(t) = P(CO) + P(H2O). Hence;

P(CO) = P(t) - P(H2O). Note that P(H2O)= 19.8 torr.

==>P(CO)= 752 torr - 19.8 torr = 732.2 torr.

STEP TWO: calculate the number of moles of Carbonmonoxide,CO.

Using the formula below;

Number of moles= pressure(P) × volume(v) / gas constant(R) × temperature (T).

That is, n= PV/RT.

n= 732 torr × 0.242 Litres/ 62.4 × 295.15.

= 9.62 × 10^-3 mol of CO.

STEP THREE:

2 moles of HCOONa = 2 moles of CO.

=> 2 moles of HCOONa = 2 moles of CO/ 2 moles of CO = 1 mol( HCOONa/ CO).

Then, 9.62 × 10^-3 mol of CO × 1 mol( HCOONa/ CO).

==> 9.62 × 10^-3 mol HCOONa × molar masss of HCOONa(68 grams/mol)

= 0.654 grams.

Therefore, the percentage of sodium formate in the original mixture = 0.654 grams/ 0.964 gram × 100 = 67.8 %.

5 0

3 years ago