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prisoha [69]
2 years ago
7

A tiny pencil mark just visible to the naked eye contains about 3 × 1017 atoms of carbon. what is the mass of this pencil mark i

n grams?
Chemistry
1 answer:
Vera_Pavlovna [14]2 years ago
6 0
<span>The number of carbon atoms in the pencil mark is 3X1017 atoms of carbon The atomic weight of carbon is 12.01 amu, so 12.01 g of carbon contains 6.022X1023 atoms. Thus equation: 12.01 g carbon/ 6.022 X 1023 atoms (3X1017 atoms) (12.01 g carbon/ 6.022X1023 atoms) - the atoms cancel out (3X1017) (12.01g carbon/6.022X1023) divide and multiply = 6x10-6g carbon The mass of this pencil mark in grams is 6x10-6g carbon</span>
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D is a correct Lewis Dot structure. Nitrogen has 4 valence electrons. 
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Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

\frac{r_{cation}}{r_{anion}}

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, Na^+ = 102 pm

Radius of cation, Br^- = 196 pm

\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

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2 years ago
The reaction of 0.779 g K with O2 forms 1.417 g potassium superoxide, a substance used in self-contained breathing devices. Dete
zhannawk [14.2K]
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Arte-miy333 [17]
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The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ΔH, ΔS, a
Nostrana [21]

The question is incomplete, the complete question is;

The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ∆H, ∆S, and ∆G for the boiling process at this temperature.

A. ∆H > 0, ∆S > 0, ∆G < 0

B. ∆H > 0, ∆S > 0, ∆G > 0

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D. ∆H < 0, ∆S > 0, ∆G > 0

E. ∆H < 0, ∆S < 0, ∆G > 0

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Secondly, a phase change from liquid to gas leads to an increase in entropy hence ∆S>0.

Thirdly, the process is spontaneous. For every spontaneous process ∆G<0

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