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AlladinOne [14]
3 years ago
7

PLZ HELP QUICKLY!!

Chemistry
1 answer:
frutty [35]3 years ago
6 0

Answer:

Robert Hooke.

Anton van Leeuwenhoek.

Matthias Schleiden.

Theodor Schwann.

Rudolf Virchow.

Explanation:

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oth hydrogen sulfide (H2S) and ammonia (NH3) have strong, unpleasant odors. Which gas has the higher effusion rate? If you opene
Ksju [112]

Answer: Odor of ammonia would we detect first on the other side of the room.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

\frac{Rate_{H_2S}}{Rate_{NH_3}}=\sqrt{\frac{M_{NH_3}}{M_{H_2S}}}

\frac{Rate_{H_2S}}{Rate_{NH_3}}=\sqrt{\frac{17.031}{34.1}}

\frac{Rate_{H_2S}}{Rate_{NH_3}}=0.71

Thus the odor of ammonia would we detect first on the other side of the room as the rate of effusion of ammonia would be faster as it has low molecular weight as compared to hydrogen sulphide.

5 0
3 years ago
Read 2 more answers
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
3 years ago
Read 2 more answers
1) If you have 2.6 moles of iron (III) oxide, how many molecules of iron (III)
Ugo [173]

Answer:

234

Explanation:

so 3 x 3 x 26 =234

8 0
3 years ago
Adam observed properties of four different waves and
cupoosta [38]

Answer:

D)The sound quality for these waves cannot be compared.

Explanation:

I've done it on e2020

5 0
4 years ago
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Calculate the Molarity when a 6.11 mL solution of 0.1 H2SO4 is diluted with 105.12 mL of water
barxatty [35]

Molarity after dilution : 0.0058 M

<h3>Further explanation </h3>

The number of moles before and after dilution is the same  

The dilution formula

 M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

M₁=0.1 M

V₁=6.11

V₂=105.12

\tt M_2=\dfrac{M_1.V_1}{V_2}=\dfrac{0.1\times 6.11}{105.12}=0.0058~M

5 0
3 years ago
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