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USPshnik [31]
2 years ago
8

Which of the following would be found ONLY in plant cells

Chemistry
1 answer:
marin [14]2 years ago
5 0

Answer:

chloroplasts, cell walls, or intracellular vacuoles

Explanation:

You didn't list the following, but I'm guessing it is chloroplasts, cell walls, or intracellular vacuoles

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A 2.5 L flask is filled with 0.25 atm SO3, 0.20 atm SO2, and 0.40 atm O2, and allowed to reach equilibrium. Assume at the temper
Anit [1.1K]

Explanation:

Reaction equation for the given chemical reaction is as follows.

      2SO_{3} \rightleftharpoons 2SO_{2} + O_{2}

Equation for reaction quotient is as follows.

         Q = \frac{P^{2}_{SO_{2}} \times P_{O_{2}}}{P^{2}_{SO_{3}}}

             = \frac{(0.20)^{2} \times 0.40}{(0.25)^{2}}

             = 0.256

As, Q > K (= 0.12)

The effect on the partial pressure of SO_{3} as equilibrium is achieved by using Q, is as follows.

  • This means that there are too much products.
  • Equilibrium will shift to the left towards reactants.
  • More SO_{3} is formed.
  • Partial pressure of SO_{3} increases.
4 0
3 years ago
Will give 20 points and brainliest help
Blababa [14]

the answer you have chosen is correct

8 0
2 years ago
How are elements arranged in a periodic table
andre [41]
Periodic table is arranged according to atomic size and other properties.
6 0
3 years ago
Read 2 more answers
Someone pls help me I will make you brain
Katena32 [7]

Answer:

it is actually b because i did this i picked b and got it right

Explanation:

4 0
2 years ago
You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli
maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

M_1V_1=M_2V_2

where, M_1 is initial molarity and  M_2 is the molarity after dilution. Similarly,  V_1 is the volume before dilution and  V_2 is the volume after dilution.

Let's plug in the values in the equation:

1.25M(363mL)=0.50M(V_2)

V_2=\frac{1.25M(363mL)}{0.50M}

V_2=907.5mL

Volume of water added = 907.5mL - 363mL  = 544.5 mL

So, 544.5 mL of water are need to be added to the original solution for dilution.

3 0
3 years ago
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