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uranmaximum [27]
3 years ago
7

What is amu of 99 % H-1, .2% H-1 and .8% H-3

Chemistry
1 answer:
ankoles [38]3 years ago
4 0

The average atomic mass of your mixture is 1.03 u .

The average atomic mass of H is the weighted average of the atomic masses of its isotopes.  

We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its % abundance).  

Thus,  

0.99    × 1.01 u = 0.998 u

0.002 × 2.01 u = 0.004 u

0.008 × 3.02 u = <u>0.024 u</u>

            TOTAL =  1.03   u

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8 0
3 years ago
The octet rule states that the formula for a compound is derived by Group of answer choices having a net charge of magnitude 8.
balandron [24]

Answer:

Atoms bonds with 8 electrons

Explanation:

Octet rule is a chemical rule that shows elements in main group bond so that each atom have 8 valence electrons in the outmost shell which make them to have similar electronic configuration with Noble gases This is important because the sharing of electrons make atoms to have full shell. Atoms with electrons not up to 8 react with more stable compound.

6 0
3 years ago
If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?
kodGreya [7K]

Answer:

[Cu^{2+}]=0.041 M

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}

[Cu^{2+}]=0.041 M

Best regards!

6 0
3 years ago
How many grams are in 9.05 x 1023 atoms of silicon?
Sergeu [11.5K]

Explanation:

Number of moles(n)=Number of atoms(N)/Avogadro's constant.

Avogadro's constant=6.02×10²³

so we have

n=9.05×10²³/6.02×10²³

n=1.0503moles.

n=mass/molar mass

1.0503=mass/28

mass=1.0503×28

mass=29.4084g

6 0
3 years ago
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8090 [49]
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7 0
3 years ago
Read 2 more answers
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