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marta [7]
1 year ago
12

The effusion rate of hcl is 43.2 cm/min in a certain effusion apparatus. what is the rate of effusion of ammonia in the same app

aratus?
Chemistry
1 answer:
Jlenok [28]1 year ago
5 0

The effusion rate is 1.125 cm/sec for ammonia.

How to find effusion rate ?

Effusion rate (r1) HCl = 43.2 cm/min

Molar mass (m2) NH3 =17.04g/mole

Molar mass (m1)  HCl    =36.46g/mole

  • Substitute the molar masses of the gases into Graham's law and solve for the ratio.
  • r1÷r2=√m2÷m1

       firstly convert 43.2 cm/min into cm/sec i.e., 0.72 cm/sec

      Then,

      0.72/r2 =√17.04/36.46

      r2= 1.125 cm/sec

Hence, the rate of diffusion of ammonia is 1.125 times faster than the rate of diffusion of hydrogen chloride.

learn more about effusion here:

brainly.com/question/2097955

#SPJ4

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Its a covalent bond for this q
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3 years ago
Which of the following solutions is acidic? [H3O+] = 1.0 x 10-10 M [H3O+] < 1.0 x 10-7 M [OH-] = 1.0 x 10-10 M [OH-] = 1.0 x
Burka [1]

Answer:

[OH-] = 1.0 x 10-10 M

Explanation:

The acidity of a solution can be determined directly from the concentration of the hydrogen ions and indirectly from the concentrations of the hydroxide ions.

Generally, for a neutral solution we have;

[H3O+] = [OH-] = 1.0 x 10-7 M

For an acidic solution;

[H3O+] > 1.0 x 10-7 M

[OH-] < 1.0 x 10-7 M

Comparing the options the correct option is;

[OH-] = 1.0 x 10-10 M

6 0
2 years ago
What is the molarity of solution obtained when 5.71 g of sodium carbonate-10-water is dissolved in water and made up to 250.0 cm
disa [49]

We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.

The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O)  is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³

The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.

Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.

5.71 grams of Na₂CO₃.10 H₂O is equal to \frac{5.71}{286}= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.

Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is \frac{0.0199 X 1000}{250} = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³


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3 years ago
4
matrenka [14]

Answer:

Yes.

Explanation:

Yes, this difference of readings will definitely affect the results of the experiment as well as the E values because the readings taken by both students are different from one another. There is a fault in one of the thermometer because both shows different readings of temperature of the same solution. This will affect the overall experiment and due to this error, we are unable to tell that which one reading is correct so the answer is uncertain or unsure.

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True

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