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snow_tiger [21]
3 years ago
11

Which observation illustrates the law of conservation of mass?

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
7 0
I think the correct answer among the choices listed above is the second option. The observation which illustrates the law of conservation of mass is that wh<span>en 3 g of carbon reacts with 8 g of oxygen, 11 g of carbon dioxide is produced. Other choices doesn't shows that mass in is equal to the mass out.</span>
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Question 1(Multiple Choice Worth 4 points)
zubka84 [21]

<u>Answer </u>

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of H_{2}O over Nine moles of O_{2}

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of Fe_{2}O_{3}.

Answer 4 : Mass of O_{2} = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

<u>Solution </u>

Solution 1 : Given,

Given mass of Fe_{2}O_{3} = 55 g

Molar mass of Fe_{2}O_{3} = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of Fe_{2}O_{3} = \frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}} = \frac{55 g}{159.69 g/mole} = 0.344 moles

Balanced chemical reaction is,

Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)

From the given reaction, we conclude that

1 mole of Fe_{2}O_{3} gives              →         3 moles of CO

0.344 moles of Fe_{2}O_{3} gives    →         3 × 0.344 moles of CO

                                                     =         1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

                    = 1.032 × 28.01

                    = 28.90 g

Solution 2 : The balanced chemical reaction is,

2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O

From the given reaction, we conclude that the Six moles of H_{2}O over Nine moles of O_{2} is the correct option.

Solution 3 : The balanced chemical reaction is,

4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}

From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of Fe_{2}O_{3}.

Solution 4 : Given,

Given mass of Zn(ClO_{3})_{2} = 150 g

Molar mass of Zn(ClO_{3})_{2} = 232.29 g/mole

Molar mass of O_{2} = 31.998 g/mole

Moles of Zn(ClO_{3})_{2} = \frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}} = (\frac{150\times 1}{232.29})moles

The balanced chemical equation is,

Zn(ClO_{3})_{2}}\rightarrow ZnCl_{2}+3O_{2}

From the given balanced equation, we conclude that

1 mole of Zn(ClO_{3})_{2} gives          →       3 moles of O_{2}

(\frac{150\times 1}{232.29})moles of Zn(ClO_{3})_{2} gives  →  [(\frac{150\times 1}{232.29})\times 3] moles of O_{2}

Mass of O_{2} = Number of moles of O_{2} × Molar mass of  O_{2} = [(\frac{150\times 1}{232.29})\times 3] \times 31.998 grams

Therefore, the mass of O_{2} = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Solution 5 : Given,

Number of moles of Na_{2}SO_{4} = 4.2 moles

Balanced chemical equation is,

H_{2}SO_{4}+2NaCN\rightarrow 2HCN+Na_{2}SO_{4}

From the given chemical reaction, we conclude that

1 mole of Na_{2}SO_{4} obtained from 2 moles of NaCN

4.2 moles of Na_{2}SO_{4} obtained   →   2 × 4.2 moles of NaCN

Therefore,

The moles of NaCN needed = 2 × 4.2 = 8.4 moles


3 0
3 years ago
Read 2 more answers
HELP QUICK ILL GIVE YOU BRAINLIEST AND 25 POINTS! bromine (br) has four energy levels. name 2 other elements that would have fou
Temka [501]

Answer:

titanium and potassium

Explanation:

6 0
3 years ago
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A resource is being used by a population.
jonny [76]

Answer:

Graph A.

Explanation:

Graph

It represents a constant function

5 0
1 year ago
Read 2 more answers
What is the mass of 0.714 moles of Mercury (I) Chloride (Hg2Cl2)?
ArbitrLikvidat [17]
Data:

m (<span>Sample Mass) = ? 
n (</span><span>Number of moles) = 0.714 mol
MM (Molar Mass) of </span>Mercury (I) Chloride (Hg_{2}  Cl_{2})
Hg = 2*200.59 = 401.18 amu
Cl = 2*35.453 = 70.906 amu
----------------------------------------
Molar Mass Hg_{2} Cl_{2} = 401.18 + 70.906 = 472.086 ≈ 472.09<span> amu or 472.09 g/mol
</span>
Formula:

n =  \frac{m}{MM}

Solving:


n = \frac{m}{MM}
0.714 =  \frac{m}{472.09}
m = 337.07226\:\to\:\boxed{\boxed{m\approx 337 g}} \end{array}}\qquad\quad\checkmark

Answer:
By approximation would be letter D) <span>337.2 g</span>


8 0
3 years ago
What is the final volume (l) of a 10.0 l system that has the pressure quartered?
mihalych1998 [28]
According to Boyle's Law, P1V1 = P2V2

where P1 and V1 are initial pressure and volume respectively. P2 and V2 are final pressure and volume receptively.

Given: P2 = 4 P1 and V1 = 10.0l

∴ V2 = 2.5 l

Answer: Final volume of system is 2.5 l


5 0
3 years ago
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