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Nataly_w [17]
3 years ago
13

In the lab, a student calculated and average P x V value of 48.0 mmHgʻL using the syringe-pressure

Chemistry
1 answer:
patriot [66]3 years ago
5 0

Answer:

0.0026 mole

Explanation:

The following data were obtained from the question:

Pressure (P) × volume (V) = P × V = 48 mmHg•L

Temperature (T) = 18.0°C.

Gas constant (R) = 62.37 mmHg•L/mol•K

Number of mole (n) =.?

Next, we shall convert 18 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 18 °C

T(K) = 18 + 273

T(K) = 291 K

Finally, we shall determine the number of mole of air in the system as shown below:

Pressure (P) × volume (V) = P × V = 48 mmHg•L

Temperature (T) = 291 K.

Gas constant (R) = 62.37 mmHg•L/mol•K

Number of mole (n) =.?

PV = nRT

48 = n × 62.37 × 291

48 = n × 18149.67

Divide both side by 18149.67

n = 48 / 18149.67

n = 0.0026 mole

Therefore, the number of mole of air in the system is 0.0026 mole

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A chemist makes of nickel(II) chloride working solution by adding distilled water to of a stock solution of nickel(II) chloride
Lady_Fox [76]

Answer:

0.0900 mol/L

Explanation:

<em>A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.</em>

Step 1: Given data

  • Initial concentration (C₁): 0.135 mol/L
  • Initial volume (V₁): 220. mL
  • Final concentration (C₂): ?
  • Final volume (V₂): 330. mL

Step 2: Calculate the concentration of the final solution

We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

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3 0
3 years ago
An alum hydrate sample was analyzed by decomposition and gave the following data:
xz_007 [3.2K]

Answer:

No, the experimental result is different from the theoretical value.

Explanation:

Based on the given information, the mass of beaker and watchglass plus alum hydrate is 102.218 grams, and the mass of beaker and watchglass is 101.286 grams. Therefore, the mass of alum hydrate is:  

= 102.218 grams - 101.286 grams

= 0.932 grams

Now the mass of anhydrous compound is,  

= 102.218 grams - 101.798 grams

= 0.42 grams

Thus, the mass of water present is,  

= 0.932 grams - 0.42 grams

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The mass percent of water is,  

= mass of water/Total mass of hydrate * 100

= 0.512 grams / 0.932 grams * 100

= 54.93 %

Hence, the experimental result in not similar to the theoretical result.  

7 0
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