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Gala2k [10]
3 years ago
10

Calculate the mass in grams of 8.35 × 1022 molecules of CBr4.

Chemistry
1 answer:
antiseptic1488 [7]3 years ago
4 0

Answer: 45.983 g CBr₄

Explanation:

To convert from moles to grams, you know that we will need molar mass and Avogadro's number.

Avogadro's number: 6.022×10²³ molecules/mol

Molar mass: 331.627 grams/mol

Now that we have what we need, you can use these to solve for grams. 8.35*10^2^2molecules CBr_{4} *\frac{1mol}{6.022*10^2^3molecules} *\frac{331.627 g}{1 mol} =45.983 g

Our final answer is 45.983 g CBr₄.

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If Log 4 (x) = 12, then log 2 (x / 4) is equal to
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The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.

<h3>What are the required properties of the logarithm?</h3>

The required logarithm properties are

logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);

Where a is the base of the logarithm.

<h3>Calculation:</h3>

It is given that,

log₄(x) = 12;

On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;

So,

log₄(x) = 12 ⇒ 4¹² = x

⇒ x = (2²)¹² = 2²⁴

Then, calculating log₂(x/4):

log₂(x/4) = log₂(2²⁴/4)

              = log₂(2²⁴/2²)

              = log₂(2²⁴ ⁻ ²)

              = log₂(2²²)

On applying the property logₐ(xⁿ) = n logₐ(x);

log₂(x/4) = 22 log₂2

We know that logₐa = 1;

So,

log₂(x/4) = 22(1)

∴ log₂(x/4) = 22.

Learn more about the properties of logarithm here:

brainly.com/question/12049968

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