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shtirl [24]
2 years ago
10

Element Y has two natural isotopes Y-63 (62.940 amu) and Y-65 (64.928 amu). Calculate the atomic mass of element Y, given the ab

undance of Y-63 is 69.17%?
Chemistry
1 answer:
djverab [1.8K]2 years ago
5 0

Answer:

Average atomic mass = 63.553 amu.

Explanation:

Given data:

Abundance of Y-63 = 69.17%

Abundance of Y-65 = 100 - 69.17 = 30.83%

Atomic mass of Y-63 = 62.940 amu

Atomic mass of Y-65 = 64.928 amu

Atomic mass of Y = ?

Solution:

Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass= (62.940×69.17)+(64.928×30.83) /100

Average atomic mass =  4353.560 + 2001.730 / 100

Average atomic mass = 6355.29 / 100

Average atomic mass = 63.553 amu.

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almond37 [142]

Answer: Option (6) is the correct answer.

Explanation:

Lattice energy is described as the energy released when formation of 1 mole of an ionic compound occurs due to the combination of its constituent ions.

Also, lattice energy is inversely proportional to the distance between the cation and anion. And, when we move down a group then there occurs an increase in the atomic radii of the atoms.

This means that smaller is the ionic distance between the cation and anion, more will be the lattice energy between the atoms.

Therefore, order from weakest to strongest lattice energy (most positive to most negative) for the given compounds is as follows.

 barium chloride < strontium chloride < calcium chloride < magnesium chloride

6 0
2 years ago
What is the stoichiometric ratio between BaCl2 and NaCl
bixtya [17]
<span>BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
7 0
2 years ago
A stock solution of Cu2+(aq) was prepared by placing 0.8875 g of solid Cu(NO3)2∙2.5 H2O in a 100.0-mL volumetric flask and dilut
Musya8 [376]

Answer:

3.816 × 10⁻³ M

Explanation:

<em>A stock solution of Cu²⁺(aq) was prepared by placing 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask and diluting to the mark with water. What is the concentration (in M) of Cu²⁺(aq)in the stock solution?</em>

<em />

We can establish the following relations:

  • The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
  • 1 mole of Cu(NO₃)₂∙2.5H₂O contains 1 mole of Cu²⁺.

The moles of Cu²⁺ in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

0.8875gCu(NO_{3})_{2}.2.5H_{2}O\times \frac{1molCu(NO_{3})_{2}.2.5H_{2}O}{232.59gCu(NO_{3})_{2}.2.5H_{2}O} \times \frac{1molCu^{2+} }{1molCu(NO_{3})_{2}.2.5H_{2}O} =3.816\times10^{-3} molCu^{2+}

The molarity of Cu²⁺ is:

\frac{3.816\times10^{-3} mol}{100.0 \times10^{-3}L} =3.816\times10^{-2}M

4 0
2 years ago
How many moles are in 987 grams of : Ra(OH)2
sergejj [24]
Your answer is 3.80 moles
8 0
2 years ago
Read 2 more answers
Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(
kkurt [141]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) 2Sr(s)+O_2(g)\rightarrow 2SrO(s)    \Delta H_1=-1184kJ

(2) SrO(s)+CO_2(g)\rightarrow SrCO_3(s)     \Delta H_2=-234kJ      ( × 2)

(3) CO_2(g)\rightarrow C(s)+O_2(g)     \Delta H_3=394kJ    ( × 2)

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ

Hence, the \Delta H^o_{rxn} for the reaction is 72 kJ.

4 0
2 years ago
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