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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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Metals of group 1 and 2 are highly electropositive and are more likely to loose electrons in a bonding situation. Therefore, the element that will most likely lose electrons to form positive ions when bonding with other elements is rubidium (Rb).

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According to Linus Pauling, electronegativity refers to the ability of an element in a compound to draw electrons towards itself.

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