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Mandarinka [93]
3 years ago
11

Which of the following equations is balanced and follows the Law of Conservation of Mass?

Chemistry
1 answer:
Jet001 [13]3 years ago
7 0

Option B is a balanced equation and follows the Law of Conservation of Mass. This is because the number of atoms of each element on both sides are the same.

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Read 2 more answers
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)

x = 1.084\times 10^{24}\,f.u.

7) The quantity of moles of potassium permanganate is:

n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }

n = 1.214\,moles

8) The number of molecules of carbon tetrachloride is:

x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

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

9) The quantity of moles of aluminium is:

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

n = 0.609\,moles

10) The number of molecules of oxygen difluoride is:

x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

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

3 0
3 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
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