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2 years ago

A gas sample occupies 5.38L at 36.0C. what is its temperature if volume is changed to 4.86L

Chemistry

1 answer:

Alina [70]2 years ago

6 0

Answer: 2.5°C

Explanation:

Initial volume V1 = 5.38 liters

Initial temperature T1 = 36.0°C

Convert temperature in Celsius to Kelvin

(32°C + 273= 305K)

Final temperature T2 = ?

Final volume V2 = 4.68 liters

According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.

Thus, Charles' Law is expressed as: V1/T1 = V2/T2

5.38/305 = 4.86/T2

To get the value of T2, cross multiply

5.38 x T2 = 4.86 x 305

5.38T2 = 1482.3

Divide both sides by 5.38

5.38T2/5.38 = 1482.3/5.38

T2 = 275.5K

Convert 275.5K to Celsius

(275.5K - 273K = 2.5°C)

Thus, the final temperature is 2.5°C

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A student determines measures the mass of one mole of carbon and finds it to be 12.22 grams. if the accepted value is 12.11 gram

Gre4nikov [31]

Answer:-

0.91% is the students % of error

Explanation: -

Accepted value= 12.11 grams

Measured value = 12.22 grams

Error = 12.22-12.11 = 0.11 grams

Percentage error = $\frac{0.11 grams}{12.11 grams}$ x100

= 0.91 %

Thus 0.91% is the students % of error

5 0

2 years ago

Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so

BartSMP [9]

Answer:

NH₃/NH₄Cl

Explanation:

We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.

$pH=pKa+log\frac{[base]}{[acid]}$

If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.

Let's consider the following buffers and their pKa.

CH₃COONa/CH3COOH (pKa = 4.74)

NH₃/NH₄Cl (pKa = 9.25)

NaOCl/HOCl (pKa = 7.49)

NaNO₂/HNO₂ (pKa = 3.35)

NaCl/HCl Not a buffer

The optimum buffer is NH₃/NH₄Cl.

4 0

2 years ago

Among the alkali metals, the tendency to react with other substances. A. does not vary among the members of the group. . B. incr

yKpoI14uk [10]

B is correct. As you move down group 1, the elements become more reactive with other elements because the electrons have a weaker attraction to their own atoms nucleus which means attraction with other elements is much stronger, making the atom more reactive.

8 0

1 year ago

You prepare a standard by weighing 10.751 mg of compound X into a 100 mL volumetric flask and making to volume. You further dilu

ArbitrLikvidat [17]

Answer:

0.12693 mg/L

Explanation:

First we calculate the concentration of compound X in the standard prior to dilution:

10.751 mg / 100 mL = 0.10751 mg/mL

Then we calculate the concentration of compound X in the standard after dilution:

0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L

Now we calculate the concentration of compound X in the sample, using the known concentration of standard and the given areas:

2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L

Finally we calculate the concentration of X in the sample prior to dilution:

0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L

4 0

1 year ago

A 50.6 grams sample of magnesium hydroxide (Mg(OH)2) is reacted with 45.0 grams of hydrochloric acid (HCl). What is the theoreti

zysi [14]

Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)

grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required

Since there are only 45.0 grams HCl, then HCl is the limiting reactant.

theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2

7 0

2 years ago

Read 2 more answers