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Valentin [98]
2 years ago
13

2. Which of the following solution contains more salt? A. 3M B. 1M C. 0.5M D. 0.25M

Chemistry
1 answer:
Tanya [424]2 years ago
5 0

Answer: 0.25M

Explanation: Because of the highest concentration of water and, if they are all the same volume, the greatest amount of water assuming it is a solid solute in a water solution.

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4. Base your answer to the following question on the elements in Group 2 on the Periodic Table. State, in terms of the number of
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Answer:

The atomic radius of strontium is larger than magnesium because there are more number of shielding electrones in strontium than magnesium, which causes electron-electron repulsion, that makes the electrons in valance shell to expand. Hence it would decrease the effective nuclear charge.

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2 years ago
How many grams are there in 12.04 x 1024 molecules of H2O2?​
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34.01468 grams

Explanation:

4 0
3 years ago
Starting with acetylene and bromoethane, show how you would use reagents from the table to synthesize 3-hexanone.
Jobisdone [24]

solution:

A = 192 x (1/2) ^ (15/5) = 192 x (1/2) ^3 = 192 x 1/8 = 24 mg

 Starting by hitting acetylene with NaNH2 to deprotonate, this C-- will attack the C connected to the Br Sn2 style to lengthen the chain by two carbons.  

Do this same thing again with the other CH of the acetylene and another bromoethaneto get a six carbon chain, namely, 3-hexyne.  

Now, reduce the alkyne to an alkene via H2/Pd/C, and that gives 3-hexene.  


3 0
3 years ago
A) How much CO is required for the production of Fe from 1000 tonnes of magnetite (Fe2O4). Assume that the iron ore is 100% pure
Kryger [21]
Its b im pretty sure
5 0
2 years ago
The radioactive substance cesium-137 has a half-life of 30 years. The amount At (in grams) of a sample of cesium-137 remaining a
stealth61 [152]

<u>Answer:</u> The amount of sample left after 20 years is 288.522 g and after 50 years is 144.26 g

<u>Explanation:</u>

We are given a function that calculates the amount of sample remaining after 't' years, which is:

A_t(t)=458\times (\frac{1}{2})^{\frac{t}{30}

  • <u>For t = 20 years</u>

Putting values in above equation:

A_t(t)=458\times (\frac{1}{2})^{\frac{20}{30}

A_t(t)=288.522g

Hence, the amount of sample left after 20 years is 288.522 g

  • <u>For t = 50 years</u>

Putting values in above equation:

A_t(t)=458\times (\frac{1}{2})^{\frac{50}{30}

A_t(t)=144.26g

Hence, the amount of sample left after 50 years is 144.26 g

6 0
3 years ago
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