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Vera_Pavlovna [14]
2 years ago
6

Which of the following experimental observations proved that the idea of matter and mass being evenly distributed in an atom was

flawed?
A. Less than 1% of the alpha particles went un-deflected through the gold foil.
B. Less than 1% of the alpha particles were deflected by angles greater than 90°.
C. The beam from the negative electrode was deflected by electrically charged rods.
D. The beam from the negative electrode cast a shadow on the positive electrode.
Chemistry
2 answers:
Ksenya-84 [330]2 years ago
7 0

A. Less than 1% of the alpha particles went un-deflected through the gold foil.

nasty-shy [4]2 years ago
6 0

Answer:

B. Less than 1% of the alpha particles were deflected by angles greater than 90°.

Explanation:

In Rutherford’s gold foil experiment most of the alpha particles pass undeflected which proved that most of the atom is empty. Less than 1% of the alpha particles were deflected by angles greater than 90°. This proved that the entire mass of the atom is confined to a small space which is highly dense.

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What is 1.23 x 10^-3 in standard notation
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Answer:

=0.00123

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The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×
Anuta_ua [19.1K]

Answer:

Kc=~1.49x10^3^4}

Explanation:

We have the reactions:

A: N_2_(_g_) + O_2_(_g_)  2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9

Our <u>target reaction</u> is:

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)

We have NO_(_g_) as a reactive in the target reaction and  NO_(_g_) is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

Then if we add reactions A and B we can obtain the target reaction, so:

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}

Kc=~1.49x10^+^3^4}

3 0
2 years ago
How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?
Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

C_{6}H_{8}O_{6}

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g

To know the percent of each element, we need to to the following:

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So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

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We know the molecular masses of each element:

C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

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Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

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