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Dafna1 [17]
3 years ago
5

Given the atomic mass of select elements, calculate the molar mass of each salt. Element Molar mass (g/mol) Beryllium (Be) 9.012

Calcium (Ca) 40.08 Copper (Cu) 63.55 Palladium (Pd) 106.42 Bromine (Br) 79.90
1. The molar mass of PdBr2 is
2. The molar mass of BeBr2 is
3. The molar mass of CuBr2 is
4. The molar mass of CaBr2 is
Chemistry
1 answer:
jeka943 years ago
6 0

Answer:

1. 266.22 g/mol

2. 168.81 g/mol

3. 223.35 g/mol

4. 199.88 g/mol

Explanation:

For you to calculate the molar mass of the salt you need to sum the molar masses of every element in the salt.

In the first salt, PdBr2_{123}, the subscript 2 means that there are 2 atoms of Br. So for you to calculate the molar mass of the salt you need to sum the molar mass of Pd and 2 times the molar mass of Br, as follows:

106 g/mol + 2(79.90 g/mol) = 266.22 g/mol

In the second salt BeBr2_{123} there are 2 atoms of Br and 1 of Be, so the molar mass is:

9.012 g/mol +2(79.90 g/mol) = 186.22 g/mol

In the third salt CuBr2_{123} there are 2 atoms of Br and 1 of Cu, so the molar mass is:

63.55 g/mol + 2(79.90 g/mol) = 223.35 g/mol

And in the fourth salt CaBr2_{123} there are 2 atoms of Br and 1 of Ca, so the molar mass is:

40.08 g/mol + 2(79.90 g/mol) = 199.88 g/mol

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Explanation:

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mass of liquid = 44.2 - 29.3

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During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
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Answer:

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Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

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it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
3 years ago
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