Which statement is true concerning energy

You only have 15g of sodium to use for this reaction. how much copper(ii must you combine with the sodium to isolate the maximum

Eva8 [605]

The balanced chemical reaction is:

CuCl2 + 2Na → 2NaCl + Cu

We are given the amount of sodium to be used up in the reaction. This will be the starting point for our calculations.

15 g Na ( 1 mol / 22.99 ) ( 1 mol Cul2 / 2 mol Na ) (134.45 g / 1 mol ) = 43.86 g CuCl2 needed to be able to obtain the maximum amount of copper.

8 0

3 years ago

A sample of methane gas occupies 4.50 gallons at a temperature of 80.0 F. If the pressure is held constant, what will be the vol

Marat540 [252]

Answer:

21.19 L or 5.6 gallons

Explanation:

Data Given

initial volume V1 = 4.50 gallons

convert volume from gallons to L

1 gallon = 3.78541 L

4. 50 gallon = 3.785 L x 4.50

So,

initial volume V1 = 17.03 L

final Volume V2 = ?

initial Temperature T1 = 80° F

convert Temperature to Kelvin

T1 = ( 80° F -32) / 1.8 +273

T1 = 26.7° C + 273 = 299.7 K

final Temperature T2 = 100° C

convert Temperature to Kelvin

T2 = 100° C + 273 = 373 K

Solution:

This problem will be solved by using charles' law equation at constant pressure.

The formula used

V1 / T1 = V2 / T2

As we have to find out volume, so rearrange the above equation

V2 = V1 x T2 / T1

Put value from the data given and conversion information in above equation

V2 = 17.03 L x 373 K / 299.7 K

V2 = 21. 19 L

So the final volume of Methane gas = 21.19 L

Or for Volume in gallons

1 L = 0.26

21.19 x 0.264 = 5.6 gallons

7 0

3 years ago

When a hurricane approaches land, people living directly on the

Tatiana [17]

Answer:

True

Explanation:

When a hurricane is dangerous enough that authorities say to evacuate, you should for your own safety.

3 0

3 years ago

Please help soon!!! About Chemistry and noble gas notations.

Svetach [21]

Answer:

play video games all days

Explanation:

ggggggggghahahahahahaha

8 0

3 years ago

A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) Minimum mass of air

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

3 years ago