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vlabodo [156]
3 years ago
8

Find the volume of 20.0 g of benzene. It's density is 0.88g/mL

Chemistry
1 answer:
damaskus [11]3 years ago
7 0
d=0,88\frac{g}{mL}\\
m=20g\\\\
d=\frac{m}{V} \ \ \ \Rightarrow \ \ \ V=\frac{m}{d}\\\\\
V=\frac{20g}{0,88\frac{g}{mL}}\approx22,73mL
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Answer:

Brittle

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Which field of science studies the composition and structure of matter?
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Q Q 3. (08.02 MC)
AnnyKZ [126]

Answer: A volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.

Explanation:

Given: V_{1} = ?,         M_{1} = 0.55 M

V_{2} = 100.0 mL,        M_{2} = 2.50 M

Formula used to calculate the volume of KBr is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.55 M \times V_{1} = 2.50 M \times 100.0 mL\\V_{1} = 455 mL

Thus, we can conclude that a volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.

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An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using t
Katena32 [7]

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

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3 years ago
What type of bond will form between potassium and iodine
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Answer:

ionic bond

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