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rosijanka [135]
4 years ago
5

What is the General equation of combustion

Chemistry
1 answer:
sdas [7]4 years ago
7 0
Combustion<span> Reactions, a </span>combustion<span> reaction always has oxygen as one reactant. The second reactant is always a hydrocarbon, which is a compound made up of carbon and hydrogen. A</span>combustion<span> reaction also always produces such as C02 and H20</span>
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19. Tryptophan is an essential _____ that can be found in the genetic code of humans.
Ivan
The answer is B tryptophan is a amino acid
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3 years ago
These cooked noodles were separated from the water they were cooked in. By which physical property were they separated? A. size
Papessa [141]

Answer:

B. Density

Explanation:

Density is the only property of liquid in the option that is feasible.

Water has no form of magnetism as it’s not a magnetic substance.

Water carries the size of the container it is put into and has no definite shape.

Water’s boiling point can only be gotten when it’s heated and it’s known to be mostly 100 degree celsius when pure and may be more or less depending on the level of impurities.

However Density is the mass per volume of water and can be calculated before the noodle is cooked which is more effective in its differentiation.

4 0
3 years ago
What are millimeters
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Millimeters are an SI unit of length that =0.0001 m
3 0
3 years ago
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which term describes an educated guess based on observation and background research of a topic? a. conclusion b. hypothesis c. l
strojnjashka [21]
It is either a or b. Hope this helps...
5 0
4 years ago
Read 2 more answers
A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant
KiRa [710]

Answer:

V_2=2.995L\\\\W=248.5J

Explanation:

Hello,

In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

P_1V_1=P_2V_2

So we solve for V_2 by firstly computing the initial pressure:

P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L}  =11.98atm

V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L

Finally, we can compute the work by using the following formula:

W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J

Best regards.

4 0
4 years ago
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