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Novosadov [1.4K]
3 years ago
11

An element has four naturally occurring isotopes with the masses and natural abundances given in the table below. Find the atomi

c mass of the element (express your answer to four significant figures and include the appropriate units).
Isotope Mass (amu) Abundance (%)
1 203.97304 1.390
2 205.97447 24.11
3 206.97590 22.09
4 207.97665 52.41
Identify the element, by spelling out its full name.
Chemistry
1 answer:
Anestetic [448]3 years ago
6 0

Answer:

Lead (Pb)

Explanation:

To find the atomic mass of the element, we need to take into consideration all the naturally occurring isotopes. We then find the atomic mass by multiplying the natural abundance of the isotopes by their mass for all of the isotopes and summing them together.

1. 1.390/100 * 203.97304 = 2.835225256

2. 24.11/100 * 205.97447 = 49.660444717

3. 22.09/100 * 206.97590 = 45.72097631

4. 52.41/100 * 207.97665 = 109.000562265

We then add all of these masses together:

109.000562265 + 45.72097631+ 49.660444717 + 2.835225256 = 207.217208548

To 4 sf = 207.2 amu

Element is Lead (Pb)

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Find the value of x in the triangle
myrzilka [38]

Answer:

55.52°

Explanation:

Concept tested: Sine rule of triangles

We need to know the sine rule

  • According to sine rule, if the three sides of a triangle are a, b and c and the corresponding angles, A, B and C
  • Then, \frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}

In this case;

  • If we take, a = 5.7 units and A = 70°, and

         b= 5 units, B = x°

  • Using the sine rule we can find the value of x

Therefore;

\frac{a}{SinA}=\frac{b}{SinB}

Then;

\frac{5.7}{Sin70}=\frac{5}{Sinx}

6.0658=\frac{5}{sinx}

Sinx=0.8243

Therefore, X = 55.52°

Therefore, the value of x in the triangle is 55.52°

6 0
2 years ago
Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixtur
yuradex [85]

<u>Answer:</u> The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For magnesium:</u>

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol

  • <u>For iron(III) chloride:</u>

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = \frac{2}{3}\times 1.708=1.114mol of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

6 0
3 years ago
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