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Serjik [45]
2 years ago
8

Which statement holds true when the molar mass of hydrogen is 2.0 grams and that of helium is 4.0 grams? Hydrogen effuses 1.4 ti

mes faster than helium. Hydrogen effuses 0.71 times faster than helium. Helium effuses 1.4 times faster than hydrogen. Helium effuses 0.71 times faster than hydrogen. The rate of effusion of both gases is the same.
Chemistry
1 answer:
SIZIF [17.4K]2 years ago
3 0
Answer:
             <span>Hydrogen effuses 1.4 times faster than Helium.

Explanation:
                   According to Graham's Law of Effusion, "the rate of effusion of a gas is inversely proportional to the square root of its density or molar mass".

Mathematically,
                                       R</span>ₐ / Rₓ  =  \sqrt{Mx/Ma}    ----- (1)
Where;
            Rₐ  =  Rate of effusion of H₂

            Rₓ  =  Rate of effusion of He

            Mₐ  =  Molar Mass of H₂  =  2 g.mol⁻¹

            Mₓ  =  Molar Mass of He  =  4 g.mol⁻¹

Putting values in equation 1,

                                       Rₐ / Rₓ  =  \sqrt{4/2}

                                       Rₐ / Rₓ  =  \sqrt{2}

                                       Rₐ / Rₓ  =  1.41

Result:
          Hence, H₂ effuses 1.41 times faster than He.
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If 133mL of a 1.2 M glucose solution is diluted to 41.0L, what is the molarity of the diluted solution?
WITCHER [35]

the formula we is as follows:-

M1V1= M2V2

where

M1=1.2

V1=0.133l

V2=41l

M2=?

1.2 × 0.133 = 41 × M2

0.1596 = 41 × M2

M2 = 0.15960/41

M2 = 0.0038926829

7 0
3 years ago
If Log 4 (x) = 12, then log 2 (x / 4) is equal to
Alexus [3.1K]

The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.

<h3>What are the required properties of the logarithm?</h3>

The required logarithm properties are

logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);

Where a is the base of the logarithm.

<h3>Calculation:</h3>

It is given that,

log₄(x) = 12;

On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;

So,

log₄(x) = 12 ⇒ 4¹² = x

⇒ x = (2²)¹² = 2²⁴

Then, calculating log₂(x/4):

log₂(x/4) = log₂(2²⁴/4)

              = log₂(2²⁴/2²)

              = log₂(2²⁴ ⁻ ²)

              = log₂(2²²)

On applying the property logₐ(xⁿ) = n logₐ(x);

log₂(x/4) = 22 log₂2

We know that logₐa = 1;

So,

log₂(x/4) = 22(1)

∴ log₂(x/4) = 22.

Learn more about the properties of logarithm here:

brainly.com/question/12049968

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8 0
1 year ago
Which of the following statements is true?
seropon [69]
Answer: option <span>D Chemical reaction rates vary with the conditions of the reaction, but nuclear decay rates do not.

Justification:

1) The rate of chemical reactions are affected by: concentration of the reactants, state of the reactants, temperature, and presence of catalizers. So the first part of the statement is true.

2) Nuclear decay rates are constant. The decay depends on the nature of the element but not the conditions. That is why dating fossils with radiactive isotopes is possible. So, the second part of the statement is true.
</span>
5 0
3 years ago
Two cars travel next to each other. One speedometer reads
Yanka [14]

Answer:

The pair of arrows which represents the relationship of speeds of the two cars is;

The second option as shown in the attached drawing

Explanation:

The given parameters are;

The reading on the speedometer of one car = 20 m/s

The reading on the speedometer of the other car = 72 km/h = 20 m/s

The blue arrow = 20 m/s

Therefore, given that the speeds of both cars are equal (20 m/s = 72 km/h = 20 m/s),  the pair of arrows that represent the relationship of speeds of the two cars is two equal length blue arrows which is the second option

The attached diagram showing the pair of arrows that represents the relationship of speeds of the two cars is drawn using Microsoft Visio.

4 0
2 years ago
A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot
kondor19780726 [428]

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

p_A=\chi_A\times P_T

p_{O_2} = partial pressure of O_2 = ?

\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}

P_{T} = total pressure of mixture  = 525 mmHg

{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles

{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles

Total moles = 1.94 + 1.35 = 3.29 moles

\chi_{O_2}=\frac{1.94}{3.29}=0.59

p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

7 0
3 years ago
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