Which of the following is NOT a reason for the experimental volume of the flask to be incorrect?

*WILL GIVE BRAINLIEST TO CORRECT ANSWER * *QUESTION IS ON PICTURE*

Ivahew [28]

Answer:

A

Explanation:

When a temperature increases particles tend to collide with each other often as they gain kinetic energy making them collide much more often and they'll collide with more energy due to the increase of kinetic energy

7 0

3 years ago

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If 5.0 liters H2 (g) at STP is heated to a temperature of 985, pressure remaining constant, the new volume of the gas will be?

vlada-n [284]

Answer:

$V_2=18 \ L \ H_2$

General Formulas and Concepts:

Chemistry - Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Charles' Law: $\frac{V_1}{T_1} =\frac{V_2}{T_2}$

Explanation:

Step 1: Define

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

Step 2: Solve for new volume

Substitute: $\frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}$
Cross-multiply: $(5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)$
Multiply: $4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K$
Isolate x: $18.0403 \ L \ H_2 = x$
Rewrite: $x=18.0403 \ L \ H_2$

Step 3: Check

We are given 2 sig figs as the smallest. Follow sig fig rules and round.

$18.0403 \ L \ H_2 \approx 18 \ L \ H_2$

5 0

3 years ago

Thallium consists of 29.5%Th-203 and 70.5%Th-205. What is the relative atomic mass of thallium?

grin007 [14]

0.295 * 203 = 53.885
0.705 * 205 = 144.525
53.885 + 144.525 = 204.41
The relative atomic mass of Thallium is 204.41

7 0

3 years ago

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Michelle is trying to find the average atomic mass of a sample of an unknown

GREYUIT [131]

The average atomic mass of her sample is 114.54 amu

Let the 1st isotope be A

Let the 2nd isotope be B

From the question given above, the following data were obtained:

Abundance of isotope A (A%) = 59.34%
Mass of isotope A = 113.6459 amu
Mass of isotope B = 115.8488 amu
Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
Average atomic mass =?

The average atomic mass of the sample can be obtained as follow:

$Average \: atomic \: mass \: = \frac{mass \: of \: A \times A\%}{100} + \frac{mass \: of \: B \times B\%}{100} \\ \\ Average \: atomic \: mass \: = \frac{113.6459\times 59.34}{100} + \frac{115.8488\times 40.66}{100} \\ \\ Average \: atomic \: mass \: = 114.54 \: amu \\ \\$

Thus, the average atomic mass of the sample is 114.54 amu

Learn more about isotope: brainly.com/question/25868336

3 0

2 years ago

John dissolves .5g of a white powder in 25g of benzene (FP 5oC) (kf benzene is 5.1) and finds the solution freezes at 3.7oC. Det

navik [9.2K]

Answer:

The compound has a molar mass of 78.4 g/mol

Explanation:

Step 1: data given

Mass of a sample = 0.5 grams

Mass of benzene = 25 grams

Freezing poing = 5 °C

Kf of benzene = 5.1 °C/m

Freezing point solution = 3.7 °C

Step 2: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 5.0 - 3.7 = 1.3 °C

⇒with i = the can't hoff factor = 1

⇒with Kf = the freezing point depression constant of benzene = 5.1 °C/m

⇒with m = the molality

1.3 = 5.1 * m

m = 1.3 / 5.1

m = 0.255 moles /kg

Step 3: Calculate moles

Molality = moles / mass benzene

0.255 molal = moles / 0.025 kg

Moles = 0.255 molal * 0.025 kg

Moles = 0.006375 moles

Step 4: Calculate molar mass of the compound

Molar mass compund = mass / moles

Molar mass compound = 0.5 grams / 0.006375 moles

Molar mass compound = 78.4 g/mol

The compound has a molar mass of 78.4 g/mol

7 0

3 years ago