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3 years ago

Which of these substances contains the most total atoms per molecule?

Chemistry

2 answers:

miss Akunina [59]3 years ago

6 0

C is your answer as it has 9 atoms

elixir [45]3 years ago

4 0

Answer: $Ni(NO_3)_2$

Explanation: A molecule is made up by combination of same atoms or different atoms.

1 molecule of $Na_2PFO_3$ consists of 2 atoms of sodium, 1 atom of phosphorous, 1 atom of flourine and three atoms of oxygen. Thus the total number of atoms is 7.

1 molecule of $H_2SO_4$ consists of 2 atoms of hydrogen, 1 atom of sulfur and 4 atoms of oxygen. Thus the total number of atoms is 7.

1 molecule of $Ni(NO_3)_2$ consists of 1 atom of nickel, 2 atoms of nitrogen and 6 atoms of oxygen. Thus the total number of atoms is 9.

1 molecule of $NbCl_5$ consists of 1 atom of niobium and 5 atoms of chlorine. Thus the total number of atoms is 6.

Thus substance which contains the most total atoms per molecule is $Ni(NO_3)_2$ .

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What is the proper way to write sodium in crystal form

Marysya12 [62]

Answer:

In the solid state Sodium has a body-centered cubic crystal structure with a = 0.428 nm and a nearest neighbor distance of 0.371 nm.

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3 years ago

PLS HELP ASAP TY. What kind of bond would many atoms of the same metal make?

alexira [117]

Answer:

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3 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .