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Alexxx [7]
4 years ago
11

Naturally occurring element X exists in three isotopic forms: X-28 (27.979 u, 77.11% abundance), X-29 (28.976 u, 8.00% abundance

), and X-30 (29.974 u, 14.89% abundance). Calculate the atomic weight of X.
Chemistry
1 answer:
slava [35]4 years ago
7 0

Answer:

The atomic weight of X is 28.356 amu

Explanation:

Step 1: Data given

X-28 has an isotopic mass of 27.979 u and an baundance of 77.11%

X-29 has an isotopic mass of 28.976 u and an baundance of 8.00 %

X-30 has an isotopic mass of 29.974 u and an baundance of 14.89 %

Step 2:  Calculate the atomic weight of X.

Atomic weight of X = 27.979 * 0.7711 + 0.28.976 * 0.08 + 29.974 *0.1489

Atomic weight of X = 28.356 amu

The atomic weight of X is 28.356 amu

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Which atom has the larger electronegativity? B, N, C, O
luda_lava [24]

Answer:

B 2.0

Explanation:

B 2.0

N 3.0

C 2.5

O 3.5

4 0
3 years ago
No. Of particles in H2SO4 in which 8grams oxygen is present????
steposvetlana [31]

Answer:

7?

Explanation:

Its somewhat hard to comprehend the question, but if the way I read it was right, its 7.

6 0
3 years ago
4g of sugar is dissolved in 46 g of water calculate the concentration as percent by mass
Lina20 [59]
The  concentration as  %  by  mass  is  calculated as below

mass  of  solute/mass of  solvent x100
mass of  solute(sugar) = 4g
mass of   solvent(water) =46  g

= 4g/  46 g x100 = 8.7%
3 0
3 years ago
Read 2 more answers
A soft drink contains 63 g of sugar in 378 g of H2O. What is the concentration of sugar in the soft drink in mass percent
Doss [256]

Answer:

\% m/m= 14.3\%

Explanation:

Hello,

In this case, the by mass percent is computed as shown below:

\% m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%

Whereas the solute is the sugar and the solvent the water, therefore, the concentration results:

\% m/m=\frac{63g}{63g+378g} *100\%\\\\\% m/m= 14.3\%

Best regards.

3 0
3 years ago
A compound found in the athletes urine had a percent composition of 80.8% carbon, 8.97% hydrogen and 10.3% oxygen. What is the E
mojhsa [17]

Answer:

C21H2802

Explanation:

C=12g/mol

H=1g/mol

O=16g/mol

Part (C) of compound-80.18%

(0.8018 x 312)/12=21

Part (H) of compound-8.97%

(0.897 x 312)/1=28

Part (O) of compund-10.3%

(0.103 x 312)/16 = 2

Therefore the emp. formula is C12H28O2

8 0
3 years ago
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