Which is formed when a standard hydrogen half-cell is connected with a salt bridge to a standard zinc half-cell?

How would a flood be a limitation for synthetic polymers that rely on natural rubber in its production?

Zepler [3.9K]

A flood, if it hits the environment of the natural rubbers, would destroy how the rubber is being produced. to have a large amount of limitation, the flood would destroy a large percentage of rubber trees. This natural rubber is needed to make synthetic polymers. Without the rubber (because of damages to it's ecosystem through the flood), there would be a limited supply, and a substancial drop on synthetic polymers.

hope this helps

3 0

3 years ago

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Which alkane may be obtained by the reduction of 2-bromo propane

Sunny_sXe [5.5K]

To reduce a haloalkane, it has to be eliminated first. This is because 2 bromopropanes are a saturated compound that can not be reduced by a saturated compound. Dehydrohalogenation of the haloalkanes into the form of propene and hydrogen bromide can lead to the elimination. The propene is then reduced to propane.

8 0

4 years ago

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____________ and _____________ are in all silicate minerals.

alina1380 [7]

I think it is B silicon and iron.

5 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

Help please!! I need to turn this in by midnight! EST

nignag [31]

Answer:

FIREWORKS : WHEN FIREWORKS EXPLODE THEY EXPEL SMOKE WHICH IS A FORM OF GAS AND THEY LIGHT UP WHICH IS ANOTHER CHEMICAL REACTION AND OF COURSE THE WHOLE FIREWORK GETS HOT ( Temperature rises )

Explanation:

you could see your reaction happens when original substance changes in color temperature formation produces light or gas in many other things

( besides it's original form! )

answer what changed from it's original form!

ex: baking a cake:

the cake change form ( rising ) due to heat ( rising temperature )

REACTION WAS:

- formation of precipitate ( solid to settle out of a solution )

- change I'm temperature to bake the cake

Extra:

you got this!

I hope I helped even just a little!

8 0

2 years ago