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tiny-mole [99]
3 years ago
15

SKJUGCiudsgcksuch clksjgc iuszskh sjgf a kuysdgf akg

Chemistry
2 answers:
taurus [48]3 years ago
8 0
<span>SKJUGCiudsgcksuch clksjgc iuszskh sjgf a kuysdgf akg</span>
Volgvan3 years ago
5 0

Answer:

sKJUGCiudsgcksuch clksjgc iuszskh sjgf a kuysdgf akg

Explanation:

fdjsk;slakjf;lkjs;lfkj ;lskj lksj lsj dflskj fldskj

:D

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kipiarov [429]

1.00

Explanation:

the density of water is always 1

7 0
3 years ago
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An aqueous potassium carbonate solution is made by dissolving 5.51 moles of K 2 CO 3 in sufficient water so that the final volum
ad-work [718]

Answer:

1.67mol/L

Explanation:

Data obtained from the question include:

Mole of solute (K2CO3) = 5.51 moles

Volume of solution = 3.30 L

Molarity =?

Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:

Molarity = mole of solute /Volume of solution

Molarity = 5.51 mol/3.30 L

Molarity = 1.67mol/L

Therefore, the molarity of K2CO3 is 1.67mol/L

3 0
3 years ago
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The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0°c is 4.48. what is the value of ka for hbro?
elena-14-01-66 [18.8K]

The solution would be like this for this specific problem:

Given:

 

pH of a 0.55 M hypobromous acid (HBrO) at 25.0 °C =  4.48

 

[H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-] <span>

Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>

 

To add, Hypobromous Acid does not require acid adjustment, which is necessary for chlorine-based product and is stable and effective in pH ranges of 5-9.<span>

</span>Hypobromous Acid combines with organic compounds to form a bromamine. Chlorine also combines with the same organic compounds to form a chloramine. <span>It is also one of the least expensive intervention antimicrobial compounds available.</span>

8 0
3 years ago
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Help asap for brainlist
Brums [2.3K]

Answer:

True po

Explanation:

Sana nakatutulong

#Carry on learning

7 0
3 years ago
HELP IM AB TO CRY
Lemur [1.5K]

Answer:

1.23 × 10³ N

Explanation:

Step 1: Given and required data

  • Mass of the person (m): 125 kg
  • Acceleration due to the gravitational force (g): 9.81 m/s²

Step 2: Calculate the force acting between the Earth and a 125-kg person standing on the surface of the Earth

We will use Newton's second law of motion.

F = m × g

F = 125 kg × 9.81 m/s²

F = 1.23 × 10³ N

8 0
3 years ago
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