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3 years ago

Is CI 3 element, compound or mixture

Chemistry

1 answer:

sattari [20]3 years ago

5 0

Answer: Element ☣

Explanation:

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3 years ago

What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances

belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope mass amu Relative abundance

1 77.9 14.4

2 81.9 14.3

3 85.9 71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9

% abundance of isotope 1 = 14.4% = $\frac{14.4}{100}=0.144$

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = $\frac{14.3}{100}=0.143$

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = $\frac{71.3}{100}=0.713$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]$

$A=84.2amu$

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

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3 years ago

What Most Often Causes The Availability Of Water To Change?

Margaret [11]

Answer:the answer is C

Explanation:

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A solution of citric acid (H₃C₆H₅O₇) with a known concentration of 0.200 M H₃C₆H₅O₇ is titrated with a 0.750 M NaOH solution. Ho

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The quantity of NaOH required to reach the third equivalence point is 20mL.

Using the titration formula,

CaVa/CbVb = Na/Nb

Where,

Ca = concentration of citric acid (0.200 M)

Cb = concentration of NaOH (0.750 M)

Va = Volume of citric acid (25.0 mL)

Vb = volume of NaOH (x mL)

Na = number of reacting mole of citric acid (3)

Nb = number of reacting mole of NaOH (1)

Therefore Vb ( x mL) =CaVaNb/CbNa

= 0.2× 25×3/0.75 ×1

= 15/0.75

Vb ( x mL) = 20 mL

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