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3 years ago

If the atom is assumed to be a sphere, what is the volume in cm3 of a single copper (cu) atom?

Chemistry

1 answer:

kompoz [17]3 years ago

5 0

Atomic radius of copper atom (r)= 128 pm

volume of sphere = 4/3 * pi * r^3 = 4/3 * 3.14 * 128^3

volume = 8780076.373 pm^3

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How many elements make up this compound C7H5(NO2)3

UkoKoshka [18]

Answer:

4 elements for C7H5(NO2)3

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3 years ago

What is the percentage of nitrogen, by mass, in calcium nitrate?.

SCORPION-xisa [38]

Answer:

<em>Nitrogen = 17.07%</em>

Explanation:

We are asked to calculate the percentage by mass of nitrogen in Ca(NO3)2.

The molar mass of calcium nitrate is 164,088 g/mol.

In this 164.088 g/mol there are 2 nitrogen atoms.

Molar mass of nitrogen = 14,0067 g/mol

% Nitrogen = molar mass of N/total molar mass of the compound

% Nitrogen =[(2 x 14.0067)/164.088]*100

% Nitrogen = 17.07%

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2 years ago

A chemical reaction produces 56.2 grams of iron (II) sulfide (FeS). How many moles of iron (II) sulfide does the reaction produc

Soloha48 [4]

Answer:

0.639

Explanation:

8 0

3 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

The source of the sun's heat is A. nuclear fission. B. nuclear fusion. C. nuclear disintegration. D. nuclear separation

ExtremeBDS [4]

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Hello!

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

❖ The source of the sun's heat is B. nuclear fusion

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

5 0

3 years ago

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