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melisa1 [442]
2 years ago
6

Which sample contains the same number of atoms as 24 grams of

Chemistry
1 answer:
Drupady [299]2 years ago
6 0

Answer:

24 g Mg

Explanation:

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PLEASE ANSWER!! I NEED HELP :)
elena-14-01-66 [18.8K]

Explanation:

To create a p-type semiconductor, group 14 semiconductors should be d o ped by atoms from group 13.

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2 years ago
What is the Hazard of Silicon?
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Silicon dust has little adverse affect on lungs and does not appear to produce significant organic disease or toxic effects when exposures are kept beneath exposure limits. Silicon may cause chronic respiratory effects. Crystalline silica (silicon dioxide) is a potent respiratory hazard.

6 0
3 years ago
What ion do acids release in solution? hydrogen hydroxide carbonate ammonium
alisha [4.7K]
Hydrogen ions or H+
8 0
3 years ago
Read 2 more answers
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
Write the empirical formula
stira [4]

Answer:

1)  NH_{4}IO_{3}\\2) Pb(IO_{3})_{4} \\3)  NH_{4}(C_{2}H_{3}O_{2})\\4) Pb(C_{2}H_{3}O_{2})_{4}

Explanation:

1) NH_{4}^{+}IO_{3}^{-} ---> NH_{4}IO_{3}\\2) Pb^{4+}(IO_{3}^{-})_{4} --->Pb(IO_{3})_{4} \\3) NH_{4}^{+}(C_{2}H_{3}O_{2})^{-} ---> NH_{4}(C_{2}H_{3}O_{2})\\4) Pb^{4+}(C_{2}H_{3}O_{2})^{-} _{4} --->Pb(C_{2}H_{3}O_{2})_{4}

7 0
3 years ago
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