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Ray Of Light [21]
2 years ago
8

Calculate the new volume in liters of a 0.500 L water bottle at 25.0 degrees when the plastic bottle is submerged in ice water a

t 1.00 degree Celsius.
Chemistry
1 answer:
ELEN [110]2 years ago
4 0

Answer:

0.02L

Explanation:

Using the formula V1/T1 = V2/T2

Given

V1 = 0.5L

T1 = 25 degrees

T2 = 1 degree

V2 =?

Substitute

0.5/25 = V2/1

Cross multiply

25V2 = 0.5

V2 = 0.5/25

V2 = 0.02L

Hence the new volume will be 0.02L

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Concentrated HCl is 37% m/m. The density of the solution is 1.19 g/mL. Assume you have 100.0 grams of SOLUTION.
sdas [7]

Answer:

M HCl sln = 12.0785 M

Explanation:

  • molarity (M) [=] mol/L
  • %mm = ((mass compound)/(mass sln))*100

∴ mass sln = 100.0 g

∴ δ sln = 1.19 g/mL

∴ % m/m = 37 %

⇒ 37 % =((mass HCl/mass sln))*100

⇒ 0.37 = mass HCl / 100.0 g

⇒ 37 g = mass HCl

∴ molar mass HCl = 36.46 g/mol

⇒ mol HCl = (37 g)*(mol/36.46 g) = 1.015 mol

⇒ volume sln = (100 g sln)*(mL/1.19 g) =  84.034 mL = 0.084034 L

⇒ M HClsln = 1.015 mol/0.084034 L

⇒ M HCl sln = 12.0785 M

7 0
2 years ago
What is the concentration of a 500 mL solution with 25 mol of HF? Write your answer with TWO decimal places and round accordingl
statuscvo [17]

Answer:

23

Explanation:

3 0
2 years ago
Which evidence cannot be found in spoiled foods?
eimsori [14]

Answer:

a. Change of state

Explanation:

Because you will see that the state has changed

8 0
1 year ago
What do butterflies and lobsters have in common?
Stella [2.4K]

Answer:

They all wear their skeletons on the outside! This is called an exoskeleton and all creatures that have their structure on the outside are included in the phylum of Arthropods.

Explanation:

3 0
3 years ago
Which of the following is the requirement for the intramolecular Sn2 reaction (intramolecular William Ether Synthesis) to occur
balandron [24]

Answer:

D. Anti-periplanar

Explanation:

In the <u>second step</u> of the intramolecular William Ether Synthesis mechanism (figure 1) we will have the attack of the negative charge of the oxygen to the carbon bond to the Br. At the same time the Br leaves, so a bond would be broken (the <u>C-Br</u> bond) and a bond would be formed (the <u>C-O</u> bond).

Now, this process can happen only if the <u>attack</u> and the <u>leaving group </u>has an anti configuration (figure 2). In an anti configuration the <u>nucleophile</u> and the <u>leaving group</u> would have <u>opposite directions</u>.

4 0
3 years ago
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