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3 years ago

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What did Bohr’s model focus on that Rutherford’s model did not?

1 answer:

wlad13 [49]3 years ago

3 0

Answer:

Because Rutherford's model was weak on the position of the electrons, Bohr focused on them. He hypothesized that electrons can move around the nucleus only at fixed distances from the nucleus based on the amount of energy they have. ... However, an electron could never exist in between two energy levels

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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

irakobra [83]

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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3 years ago

In an Honors organic chemistry lab, a student devised an experiment in which she would treat benzoic acid with t-butanol in an a

oksian1 [2.3K]

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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3 years ago

The atomic number of a metal is...<br> A. 10<br> B. 11<br> C. 14 <br> D. 20

Lilit [14]

Your closes answer would be a.10

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3 years ago

Read 2 more answers

Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y

olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

$A=A_o\times e^{-\lambda t}$

Where:

$\lambda$ = decay constant

$A_o$ =concentration left after time t

$t_{\frac{1}{2}}$ = Half life of the sample

Half life of Pu-239 = $t_{\frac{1}{2}}=24,000 y$ [

$\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]$

Let us say amount present of Pu-239 today = $A_o=x$

A = ?

$A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}$

$A=0.9715\times x$

$\frac{A}{A_0}=\frac{A}{x}=0.9715$

0.9715 Fraction of Pu-239 will be remain after 1000 years.

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3 years ago

1. Decomposition reactions can be classified into three types. Pick the correct

Alex777 [14]

Answer:

Explanation:

Option A is the correct answer

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