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Hunter-Best [27]
2 years ago
10

I need to know what’s the error in each equation help !!!

Chemistry
1 answer:
lorasvet [3.4K]2 years ago
7 0
Good luck my guy your going to need it
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What is the mass, in grams, of 1.33 mol of water, H2O?
Len [333]

Answer:

24.0g H2O

Explanation:

1.33 mol (18.016g/1 mol) = 24.0g H2O

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3 years ago
The properties of a substance are not affected by chemical reactions.<br> O True<br> O False
ElenaW [278]

Answer:

False

Explanation:

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Tails of phospholipids that don't like water
densk [106]
Those tails are called hydrophobic. You can note the etymology: hydro= water, phobi = fear, aversion, dislike.

Phospholipds' tail is a long non polar chain, made of Carbon and Hydorgens, that rejects water (a polar solvent) and is attracted to non-polar compounds (oil for example).  That is why that tails can atract dirt.
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If gas is $4.00/gal, how much does it cost to travel from Portland, Maine, to Orlando, Florida (a distance of 1,500 mi)? Give yo
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it cost $200 , the answer is not $6000

6 0
3 years ago
Read 2 more answers
If 57.3 l of 0.497 m koh is required to completely neutralize 39.5 l of a CH3COOH solution. What is the molarity of the acetic a
bearhunter [10]

The molarity of the solution will be 0.72 m.

The majority of reactions take place in solutions, making it crucial to comprehend how the substance's concentration is expressed in a solution when it is present. The number of chemicals in a solution can be stated in a variety of ways, including.

The symbol for it is M, and it serves as one of the most often used concentration units. Its definition states how many moles of solute there are in a liter of solution.

Given data:

V_{1} =57.3 L\\V_{2} = 39.5 L\\M_{1} = 0.497 m\\\\M_{2} = ?

Molarity can be determined by the formula:

M_{1} V_{1} = M_{2} V_{2}

where, M is molarity and V is volume.

Put the value of given data in above equation.

57.3 × 0.497 m = M × 39.5 L

M = 0.72 m

Therefore, the molarity of the solution will be 0.72 m

To know more about molarity

brainly.com/question/18648803

#SPJ4

6 0
1 year ago
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