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SashulF [63]
3 years ago
8

What does this equation describe?

Chemistry
1 answer:
Gwar [14]3 years ago
5 0

Answer:

the energy of photons emitted by electron dropping from an excited energy state

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A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal
bixtya [17]

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V =  Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

8 0
3 years ago
Explain why mixing of red paint with white paint does not constitute a chemical reaction even though the product has a different
MArishka [77]

Answer:

Explanation:

In a chemical change through chemical reactions, a new kind of matter forms. When a paint is mixed together, a physical change has occurred and not a chemical change. The different color one perceives is just a mere overlap between the reflected white and red color which produces another hue.

Not all color changes infers a chemical reaction. A paint is a complex mixture on its own.

5 0
3 years ago
Solve the following problem and round the correct number of significant
RUDIKE [14]

Answer:

4.5 + 2.34 = 6.8

Sig Figs:

2

Decimals:

1

Scientific Notation:

6.8 × 10ⁿ

Words:

six point eight

5 0
3 years ago
Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M
kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

3 0
3 years ago
A chemist has to prepare 250.0 mL of a 0.300 M Na2SO4(aq) solution. What mass, in grams, of sodium sulfate (formula mass 142.05
Sedbober [7]

The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g

<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
  • Volume = 250 mL = 250 / 1000 = 0.25 L
  • Molarity = 0.3 M
  • Mole of Na₂SO₄ =?

Mole = Molarity x Volume

Mole of Na₂SO₄ = 0.3 × 0.25

Mole of Na₂SO₄ = 0.075 mole

<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
  • Molar mass of Na₂SO₄ = 142.05 g/mol
  • Mole of Na₂SO₄ = 0.075 mole
  • Mass of Na₂SO₄ =?

Mass = mole × molar mass

Mass of Na₂SO₄ = 0.075 × 142.05

Mass of Na₂SO₄ = 10.65 g

Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.

Learn more about molarity:

brainly.com/question/15370276

6 0
1 year ago
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