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DochEvi [55]
3 years ago
5

Fictitious element X has an a average atomic mass of 254.9 and only 2 isotopes, One of its isotopes has an abundance of 72.00 an

d a mass of 250.9 amu. What is the atomic mass of the second isotope?
265.2

20.80

245.9

251.9

none of the above

all of the above
Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
8 0

Answer:

265.2amu

Explanation:

Given parameters:

Atomic mass  = 254.9amu

Abundance of isotope 1 = 72%

Atomic mass of isotope 1  = 250.9amu

Abundance of isotope 2  = 100  - 72  = 28%

Unknown:

Atomic mass of isotope 2 = ?

Solution:

To find the atomic mass of isotope 2, use the expression below:

Atomic mass = (abundance of isotope 1 x atomic mass of isotope 1) + (abundance of isotope 2 x atomic mass of isotope 2)

 Now insert the parameters and find the unknown;

  254.9  = (0.72 x 250.9)  + (0.28 x Atomic mass of isotope 2)

 254.9  = 180.648 + 0.28x atomic mass of isotope 2

 254.9 - 180.648  = 0.28x atomic mass of isotope 2

    74.25  = 0.28 x atomic mass of isotope 2

    Atomic mass of isotope 2 = 265.2amu

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2 mole MnO₂

Explanation:

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4 years ago
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How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.
MrRa [10]
<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

   N₂(g) + 3H₂(g) → 2NH₃(g)

<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

                                  = 2.3 moles × 2

                                  = 4.6 moles

<h3>Step 3: Calculating the mass of ammonia produced </h3>

Mass = Moles × molar mass

Molar mass of ammonia gas = 17.031 g/mol

Therefore;

Mass = 4.6 moles × 17.031 g/mol

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Thus, the mass of NH₃ produced is 78.34 g

3 0
3 years ago
What is the concentration of an unknown Mg(OH)2 solution if it took an average of 15.4mL of
vova2212 [387]

Answer:

0.077M

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2HCl + Mg(OH)2 —> MgCl2 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 2

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

Concentration of base Cb =...?

Volume of base (Vb) = 10mL

Concentration of acid (Ca) = 0.1M

Volume of acid (Va) = 15.4mL

Step 3:

Determination of the concentration of the base, Mg(OH)2.

The concentration of the base can be obtained as follow:

CaVa/CbVb = nA/nB

0.1 x 15.4 /Cb x 10 = 2/1

Cross multiply to express in linear form

Cb x 10 x 2 = 0.1 x 15.4

Divide both side by 10 x 2

Cb = (0.1 x 15.4) /(10 x 2)

Cb = 0.077M

Therefore, the concentration of the base, Mg(OH)2 is 0.077M

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