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Alla [95]
3 years ago
10

X

Chemistry
2 answers:
Butoxors [25]3 years ago
6 0
A. Burning a match because it changes its form
Marina CMI [18]3 years ago
3 0
A. burning a match !!!!
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The diagram shows an animal cell.
LenKa [72]

Answer:

X  

Explanation:

Assume your diagram is like the one below.

X represents a mitochondrion.

That's where the Tricarboxylic Acid Cycle converts a single glucose molecule into six molecules of CO₂.

W is wrong. It represents a vacuole, which can store both nutrients and waste products for later elimination.

Y is wrong. It represents the nucleolus, which plays a critical role in the synthesis of ribosomes.  

Z is wrong. It represents the cytoplasm, which is where cell processes like glycolysis and protein synthesis take place.

8 0
3 years ago
Which of the following is in the formula for an acid?<br><br> H<br> OH-<br> hydroxide<br> oxide
avanturin [10]
A.) "Hydrogen" <span>is in the formula for an acid

Hope this helps!</span>
8 0
3 years ago
Read 2 more answers
Yes thank you Great explanation
stiv31 [10]

Answer:

you're welcome

5 0
3 years ago
Which factor is a measure of the oxygen in a body of water?
KATRIN_1 [288]
B. Dissolved Oxygen

is a measure of the oxygen in a body of water.
6 0
3 years ago
calculate the freezing point of 3.46 gram of a compound X in 160 gram of benzene when a separate sample of X was vaporized it's
Temka [501]

Answer: The freezing point of 3.46 gram of a compound X in 160 gram of benzene is 4.38^0C

Explanation:

The relation of density and molar mass is:

d=\frac{PM}{RT}

where

d = density = 3.27 g/ L

P = pressure of the gas  = 773 torr = 1.02 atm   (760 torr = 1atm)

M = molar mass of the gas  = ?

T = temperature of the gas = 116^0C=(116+273)K=389K

R = gas constant  = 0.0821Latm/Kmol

M=\frac{dRT}{P}=\frac{3.27g/L\times 0.0821Latm/Kmol\times 389K}{1.02atm}=102.3g/mol

The relation of depression in freezing point with molality:

\Delta T_f=k_f\times m

\Delta T_f = depression in freezing point = T_f^0-T_f = 5.45-T_f

k_f = freezing point constant  = 5.1

m = molality = \frac{\text {moles of X}}{\text {weight of solvent in kg}}=\frac{3.46\times 1000}{102.3\times 160}=0.21

5.45-T_f=5.1\times 0.21

T_f=4.38^0C

Thus the freezing point of 3.46 gram of a compound X in 160 gram of benzene is 4.38^0C

5 0
3 years ago
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