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Serggg [28]
3 years ago
12

F 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced? (Hint: use what you know about limiting

reactant) *
10 points
Chemistry
1 answer:
Mars2501 [29]3 years ago
4 0

Answer:

1.5 mol H2O

Explanation:

2C2H6 + 7O2 -> 4CO2 + 6H2O

The limiting reactant is C2H6

30 g C2H6 ->1 mol C2H6

15 g C2H6   -> x                      x= 0.5 mol C2H6

2 mol C2H6 -> 6 mol H2O

0.5 mol c2H6 -> x                    x = 1.5 mol H2O

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The store paid 4.50 for a book and sold it for $7.65.what is the profit as a percent of the cost to the store? A. 3. 15% b. 58.8
xeze [42]

Answer: This would be C.) 70%

Explanation:

4 0
2 years ago
Based on the vibrational difference you observed, state a hypothesis to explain why diatomic molecules in the air, such as oxyge
mihalych1998 [28]

Answer:

See explanation

Explanation:

The ability of a gas to function as a green house gas depends on its ability to absorb infra red rays. In turn, the absorption of infrared red rays depends on whether or not the molecule is IR active.

The triatomic molecules such as methane and water are IR active. Only IR active molecules can lead to green house effect.

Note that for a molecular vibrational mode to be IR active, the dipole moment of the molecule is changed as the vibration occurs .

5 0
2 years ago
Match each vocabulary word with the correct definition.
Nikolay [14]
1. antibody
2. antigen
3. leukocyte
4. phagocytosis
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4 0
3 years ago
Read 2 more answers
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
Niobium-91 has a half-life of 680 years. After 2,040 years, how much niobium-91 will remain from a 300.0-g sample? 3 g 18.75 g 3
Vadim26 [7]
Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g
7 0
3 years ago
Read 2 more answers
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