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velikii [3]
3 years ago
5

It takes 167 s for an unknown gas to effuse through a porous wall and 99 s for the same volume of n2 gas to effuse at the same t

emperature and pressure. what is the molar mass of the unknown gas?
Chemistry
1 answer:
irakobra [83]3 years ago
3 0

Answer : The molar mass of the unknown gas will be 79.7 g/mol


Explanation : To solve this question we can use graham's law;


Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1

We can compare the rates of both the gases;


So here, Rate of gas 2 / Rate of gas 1 = \sqrt{(molar mass 1 / molar mass 2)}

Now, 1.687 = square root [\sqrt{(molar mass 1) / (28.01 g/mol N_{2})} ]


When we square both the sides we get;


2.845 = (molar mass 1) / (28.01 g/mol N2)


On rearranging, we get,


2.845 X (28.01 g/mol N2) = Molar mass 1

So the molar mass of unknown gas will be = 79.7 g/mol

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Complete the table for ion charge based upon their losing or gaining electrons in the outer shell. (Use the periodic table as ne
PolarNik [594]

Answer:

Explanation:

Group one:

The elements of group one shows +1 charge because these all are metals and lose their one valance electron.

Hydrogen lithium sodium potassium rubidium cesium francium

Group 2:

The elements of group two shows +2 charge because these all alkali metals and lose their two valance electrons.

beryllium magnesium calcium strontium barium radium

Group 3:

The elements of group three-B shoes +3 charge by losing three valance electrons.

Scandium yttrium lanthanum actinium

Group 4:

The elements of group 4th A and 4th B lose four electrons or gain four electrons to complete the octet and shows +4 or -4 charge.

Group 5:

Group 5th elements gain three electrons and shows -3 charge to complete the 8 electrons. (octet).

It involve the elements of group 5th A.

Group 6:

The elements of group 6A gain two electrons to complete the octet and shows -2 charge.

Group 7:

The elements of group 7A gain one electron to complete the octet and shows -1 charge.

Group 8:

The elements of group 8A are noble gases and have complete octet. That's why shows 0 charge.

4 0
3 years ago
Is this chemical equation balanced?
cluponka [151]

Answer:

Yes

Explanation:

5 0
3 years ago
Read 2 more answers
True or false
sweet-ann [11.9K]

Answer:

True.

But it only changes in physical change.

How?

Explanation:

The chemical reaction produces a new substance with new and different physical and chemical properties. Matter is never destroyed or created in chemical reactions. The particles of one substance are rearranged to form a new substance.

In a physical change, a substance's physical properties may change.

A chemical change is a permanent change. A Physical change affects only physical properties i.e. shape, size, etc. ... Some examples of physical change are freezing of water, melting of wax, boiling of water, etc. A few examples of chemical change are digestion of food, burning of coal, rusting, etc.

Hope this helps!

8 0
3 years ago
Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80
Damm [24]

Answer:

4.8 %

Explanation:

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³      (1l= 1000 cm³)

                            = 1010 g

% (by mass) = 48.00 g/ 1010 g  x 100 = 4.8 %

6 0
3 years ago
Consider a solution that is 1.3×10−2 M in Ba2+ and 2.0×10−2 M in Ca2+. Ksp(BaSO4)=1.07×10−10 Ksp(CaSO4)=7.10×10−5 If sodium sulf
skad [1K]

Answer:

Explanation:

Ksp(BaSO4)=1.07×10−10

BaSO₄ → Ba²⁺ + SO₄²⁻

1.07×10⁻¹⁰ = ( Ba²⁺) × ( SO₄²⁻)

but Ba²⁺ = 1.3×10⁻² M

1.07×10⁻¹⁰  = 1.3×10⁻² M × ( SO₄²⁻)

( SO₄²⁻)  = 1.07×10⁻¹⁰  / 1.3×10⁻² = 0.823 × 10⁻⁸ M

while Ksp(CaSO4)=7.10×10−5

CaSO₄ → Ca²⁺ + SO₄²⁻

7.10×10⁻⁵ = 2.0×10⁻² × ( SO₄²⁻)

( SO₄²⁻)  = 7.10×10⁻⁵  /  2.0×10⁻² = 3.55 × 10⁻³ M

comparing the concentration of sulfate ions, Ba²⁺ cation will precipitate first because the Ba²⁺ requires 0.823 × 10⁻⁸ M sodium sulfate which less compared the about needed by CaSO₄

7 0
3 years ago
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