Experiments<br> 11. What 2 factors increase the validity of a scientific experiment?

9. Suppose that 25.0 mL of a gas at 725 mm Hg and 20°C is converted to standard

storchak [24]

Answer:

V₂ = 22.23 mL

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 25 mL

Initial pressure = 725 mmHg (725/760 =0.954 atm)

Initial temperature = 20 °C (20 +273 = 293 K)

Final pressure = standard = 1 atm

Final temperature = standard = 273.15 K

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 0.954 atm × 25 mL × 273.15 K / 293 K × 1 atm

V₂ = 6514.63 mL . atm . K / 293 K . atm

V₂ = 22.23 mL

8 0

2 years ago

An electron is present at an energy level with an energy of 16.32 x 10−19 J. During a transition to another level, 5.4 x 10−19 J

kolbaska11 [484]

Answer:

D. The electron moved up to an energy level and has an energy of 21.72 x 10−19 J.

3 0

2 years ago

An unknown compound contains only C, H, and O. Combustion of 8.50 g of this compound produced 20.0 g CO2 and 5.46 g H2O. What is

Galina-37 [17]

Answer:

The answer to your question is: C₃H₄O

Explanation:

Data

CxHyOz = 8.5 g

CO2 = 20 g

H2O = 5.46 g

Reaction

CxHyOz + O2 ⇒ CO2 + H2O

CO2

MW = 44g

44g CO2 ----------------- 12g of C

20g CO2 ---------------- x

x = (20 x 12) / 44

x = 5.45 g of C

# of moles = n = 5.45 / 12 = 0.454 mol of C

H2O

MW = 18 g

18 g H2O ------------------- 2g of H

5.46 g -------------------- x

x = (5.46 x 2) / 18 = 0.61 g of H

n = 0.61 / 1 = 0.61 moles of H2

Mass of O2

mass CxHyOz = mass CO2 + mass H2 + mass O2

mass O2 = 8.5 - 5.45 - 0.61

mass O2 = 2.44g

n = 2.44 / 16 = 0.153 mol of O2

Now, divide by the lowest number of moles

0.454 mol of C/ 0.153 = 2.97 ≈ 3

0.61 moles of H2/ 0.153 = 3.99 ≈ 4

0.153 mol of O2/ 0.153 = 1

Then, the empirical formula is: C₃H₄O

7 0

3 years ago

If a substance is a bad conductor as a solid, but a good conductor as a liquid or when dissolved into water, which type of bond

GenaCL600 [577]

An ionic bond. covalent bonds do not conduct in a solid or a liquid state, ionic bonds conduct in a liquid state and not in a solid state.

6 0

2 years ago

A student performed a serial dilution on a stock solution of 1.33 M NaOH. A 1.0 mL aliquot of the stock NaOH (ms) was added to 9

serg [7]

Answer:

The correct answer is 1.33 x 10⁻⁵ M

Explanation:

The concentration of the stock solution is: C= 1.33 M

In the first dilution, the student added 1 ml of stock solution to 9 ml of water. The total volume of the solution is 1 ml + 9 ml = 10 ml. So, the first diluted concentration is:

C₁= 1.33 M x 1 ml/10 ml = 1.33 M x 1/10 = 0.133 M

The second dilution is performed on C₁. The student added 1 ml of 0.133 M solution to 9 ml of water. Again, the total volume is 1 ml + 9 ml = 10 ml. The second diluted concentration is:

C₂= 0.133 M x 1 ml/10 ml = 0.133 M x 1/10= 0.0133 M

Since the student repeated the same dilution process 3 times more (for a total of 5 times), we have to multiply 5 times the initial concentration by the factor 1/10:

Final concentration = initial concentration x 1/10 x 1/10 x 1/10 x 1/10 x 1/10

= initial concentration x (1/10)⁵

= 1.33 M x 1 x 10⁻⁵

= 1.33 x 10⁻⁵ M

3 0

3 years ago

Experiments 11. What 2 factors increase the validity of a scientific experiment?