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krok68 [10]
3 years ago
11

I'm super lost on beta, alpha, and gamma decay.

Chemistry
1 answer:
astra-53 [7]3 years ago
5 0

Answer:

then,

Explanation:

anong meron,hahahh

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Read 2 more answers
What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?
Law Incorporation [45]
The answer is 0.975 L

Volume = mol/Molarity

We have molarity (0.788 M) and we need mol and volume. Let's first calculate number of moles of CaCl2 in 85.3 g:

Molar mass of CaCl2 is sum of atomic masses of Ca and Cl:
Mr(CaCl2) = Ar(Ca) + 2Ar(Cl) = 40 + 2 * 35.45 = 40 + 70.9 = 110.9 g/mol

So, if 110.9 g are in 1 mol, 85.3 g will be in x mol:
110.9 g : 1 mole = 85.3 g : x
x = 85.3 g * 1 mole / 110.9
x = 0.769 moles

Now, calculate the volume:
V = 0.769/0.788
V = 0.975 L
5 0
3 years ago
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