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S_A_V [24]
2 years ago
7

A hypothetical element, E, has two stable isotopes. One isotope has a natural abundance of 68.037% and has an atomic mass of 46.

449 u. If the atomic weight of E is 47.574 u, what is atomic mass (in units of u) of the second isotope.
Chemistry
1 answer:
I am Lyosha [343]2 years ago
6 0
  • If the abundance of the first isotope is 68.037%, then the abundance of the second isotope is 100%-68.037%.

Substituting into the atomic mass formula,

47.574=(46.449)(0.68037)+x(1-0.68037)\\\\ 47.574=31.60250613+0.31963x\\\\15.97149387=0.31963x\\\\x \approx \boxed{49.969 \text{ u}}

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One mole of a metallic oxide reacts with one mole of hydrogen to produce two moles of the pure metal
Rudiy27

Answer:

Lithium oxide, Li₂O.

Explanation:

Hello!

In this case, according to the given amounts, it is possible to write down the chemical reaction as shown below:

M_2O+H_2 \rightarrow 2M+H_2O

Which means that the metallic oxide has the following formula: M₂O. Next, we can set up the following proportional factors according to the chemical reaction:

5.00gM_2O*\frac{1molM_2O}{(2x+16)gM_2O}*\frac{2molM}{1molM_2O}*\frac{xgM}{1molM}   = 2.32gM

Thus, we perform the operations in order to obtain:

\frac{10x}{2x+16}=2.32

So we solve for x as shown below:

10x = 2.32(2x+16)\\\\10x = 4.64x+37.12\\\\x = \frac{37.12}{10-4.64}\\\\x= 6.93 g/mol

Whose molar mass corresponds to lithium, and therefore, the metallic oxide is lithium oxide, Li₂O.

Best regards!

8 0
3 years ago
The isotope Np-238 has a half life of 2.0 days if 96 grams of it were present on Monday how much will remain six days later
Kipish [7]

Answer:

12.02 g

Explanation:

From the question given above, the following data were obtained:

Half life (t½) = 2 days

Original amount (N₀) = 96 g

Time (t) = 6 days

Amount remaining (N) =..?

Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:

Half life (t½) = 2 days

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 2

K = 0.3465 /day

Therefore, the rate of disintegration of the isotope is 0.3465 /day.

Finally, we shall determine the amount of the isotope remaining after 6 days as follow:

Original amount (N₀) = 96 g

Time (t) = 6 days

Decay constant (K) = 0.3465 /day.

Amount remaining (N) =.?

Log (N₀/N) = kt / 2.303

Log (96/N) = (0.3465 × 6) / 2.303

Log (96/N) = 2.079/2.303

Log (96/N) = 0.9027

Take the anti log of 0.9027

96/N = anti log (0.9027)

96/N = 7.99

Cross multiply

96 = N × 7.99

Divide both side by 7.99

N = 96 /7.99

N = 12.02 g

Therefore, the amount of the isotope remaining after 6 days is 12.02 g

3 0
2 years ago
How many moles are in 1.20 times 10^25 atoms of phosphorus
Stels [109]

19.927 moles are in 1.20 times 10^{25} atoms of phosphorus.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 ×10^{23} of some chemical unit, be it atoms, molecules, ions, or others.

1 mole of any substance contain Avogadro's number of molecules so we can calculate the number of moles by dividing the provided number of atoms over Avogadro's number to obtain the number of moles .

Moles= \frac{Atoms}{\;Avogadro's \;number }

Moles=  1.20 X 10^{25} atoms ÷ 6.022 X 10^{23}

= 19.927

Hence, 19.927 moles are in 1.20 times 10^{25} atoms of phosphorus.

Learn more about moles here:

brainly.com/question/26416088

#SPJ1

5 0
1 year ago
Whats the molar mass of CO
g100num [7]
28.01 g/mol

hope that helped
7 0
3 years ago
Read 2 more answers
The amount of 217 mg of an isotope is given by A(t) = 217 € -0.0171, where t is time in years since the initial amount of 217 mg
Maru [420]

The amount left after 20 years = 154.15 mg

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually radioactive elements have an unstable atomic nucleus.  

The main particles are emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles  

The decay formula for isotope :

\tt \large{\boxed{\bold{A(t)=217e^{-0.0171t}}}

Then for t=20 years, the amount left :

\tt A(t)=217e^{-0.0171\times 20}\\\\A(t)=154.15~mg

4 0
3 years ago
Read 2 more answers
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