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Airida [17]
3 years ago
8

What is the word equation of 1 H2 + 1 Cl2 → 2 HCI

Chemistry
1 answer:
Tom [10]3 years ago
5 0

Answer:

1 dihydrogen reacts with 1 chlorine to form two hydrogen chloride (edit: hydrogen chloride is also referred to as hydrochloric acid)

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How would the concentration of silver ions compare in a 1.0 x 10-16 L saturated solution to a 1.5 x 10-16 L saturated solution?
VMariaS [17]
The second one is more concentrated as they both times with the same thing but the second one (1.5) is bigger
3 0
3 years ago
Why does the Moon orbit Earth instead of the Sun?
Vera_Pavlovna [14]
Gravity depends on distance and the moon is closer to earth
4 0
3 years ago
The addition of phosphate to a chemical compound is called A. glycolysis. B. oxidation. C. reduction. D. phosphorylation.
stich3 [128]

Answer:

Option (D)

Explanation:

Phosphorylation can be simply defined as the addition of a phosphate group to an organic and inorganic molecule. This process helps in regulating the processes that occur in the cells. It leads to the growth and development of cells and this process is efficiently carried out with the help of enzymes like kinase. It also plays an important role in transferring the signals within the cells, synthesis, and functioning of proteins within the cells, and storing as well as releasing of energy.

Thus, the correct answer is option (D).

5 0
3 years ago
Which of the following is NOT a characteristic of a compound?
MissTica

Answer:

I think it's D

sorry if I'm wrong.

Explanation:

4 0
3 years ago
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
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