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Tomtit [17]
2 years ago
13

A sample of strontium bicarbonate weighs 5.0020 mg. How many oxygen atoms are in the sample? I have the answer, but don't unders

tand how to get there.
Chemistry
1 answer:
hodyreva [135]2 years ago
5 0

There are 14.4 * 10^18 oxygen atoms in 5.0020 mg of  strontium bicarbonate.

Mass of strontium bicarbonate Sr(HCO3)2 = 5.0020 mg

Molar mass of strontium bicarbonate Sr(HCO3)2 = 209.6537 g/mol

Number of moles of strontium bicarbonate Sr(HCO3)2 = 5.0020 * 10^-3g/209.6537 g/mol

= 2.3858 * 10^-5 moles

Given that we have 6 oxygen atoms per molecule of Sr(HCO3)2, the total number of oxygen atoms in Sr(HCO3)2 becomes;

2.3858 * 10^-5 moles * 6.02 * 10^23 = 14.4 * 10^18 oxygen atoms

Learn more: brainly.com/question/9743981

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What happen when you add heat to solids and liquids?​
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When a mercury-202 nucleus is bombarded with a neutron, a proton is ejected. What element is formed?
Galina-37 [17]

That will make a gold-202 nucleus.

<h3>Explanation</h3>

Refer to a periodic table. The atomic number of mercury Hg is 80.

Step One: Bombard the \displaystyle ^{202}_{\phantom{2}80}\text{Hg} with a neutron ^{1}_{0}n. The neutron will add 1 to the mass number 202 of ^{202}_{\phantom{2}80}\text{Hg}. However, the atomic number will stay the same.

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^{202}_{\phantom{2}80}\text{Hg} + ^{1}_{0}n \to ^{203}_{\phantom{2}80}\text{Hg}.

Double check the equation:

  • Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
  • Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.

Step Two: The ^{203}_{\phantom{2}80}\text{Hg} nucleus loses a proton ^{1}_{1}p. Both the mass number 203 and the atomic number will decrease by 1.

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Refer to a periodic table. What's the element with atomic number 79? Gold Au.

^{203}_{\phantom{2}80}\text{Hg} \to ^{202}_{\phantom{2}79}\text{Au} + ^{1}_{1}p.

Double check the equation:

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A gold-202 nucleus is formed.

6 0
3 years ago
Unit: Chemical Quantities
Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 1.993\,moles

2) The number of atoms of helium is:

x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)

x = 9.033\times 10^{23}\,atoms

3) The quantity of moles of carbon monoxide is:

n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.689\,moles

4) The number of molecules of sulfur dioxide is:

x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.355\times 10^{24}\,molecules

5) The quantity of moles of sodium chloride is:

n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.399\,moles

6) The number of formula units of magnesium iodide is:

x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)

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n = 1.214\,moles

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x = 1.506\times 10^{23}\,molecules

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n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

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x = 2.120\times 10^{24}\,molecules

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