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marin [14]
3 years ago
6

Please help ill give you points and brainliest please ​

Chemistry
1 answer:
Paul [167]3 years ago
3 0

Answer:

Answers with detail are given below

Explanation:

1) Given data:

Mass of Rb₃Rn = 76.19 g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 478.43 g/mol

Number of moles = 76.19 g/ 478.43 g/mol

Number of moles = 0.16 mol

2) Given data:

Mass of FrBi₂ = 120.02 g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 640.96 g/mol

Number of moles = 120.02 g/640.96 g/mol

Number of moles = 0.19 mol

3) Given data:

Mass of Zn₂F₃ = 88.24 g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 187.73 g/mol

Number of moles = 88.24 g/ 187.73 g/mol

Number of moles = 0.47 mol

4) Given data:

Number of moles of Sb₄Cl = 1.20 mol

Mass of Sb₄Cl = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 522.49 g/mol

Mass = Number of moles × molar mass

Mass = 1.20 mol ×  522.49 g/mol

Mass = 626.99 g

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prisoha [69]

Answer:

Question 4 is- Solubility

Question 5 is- Suspension

Hopes this helps >:D

6 0
3 years ago
Today, nuclear power plants rely on fission. While fusion reactions have been used in nuclear bombs, many scientists and enginee
Zigmanuir [339]

Explanation:

nuclear fusion yields more energy than nuclear fission and the products of the reaction are not radioactive

5 0
3 years ago
Number of grams of hcl that can react with 0.500 grams of Al(OH)3
vlabodo [156]

Answer:

0.7g of HCl

Explanation:

First, let us write a balanced equation for the reaction between HCl and Al(OH)3.

This is illustrated below:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:

Molar Mass of Al(OH)3 = 27 + 3(16+1)

= 27 + 3(17) = 27 + 51 = 78g/mol.

Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g

Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

From the equation above,

78g of Al(OH)3 reacted with 109.5g of HCl.

Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl

7 0
3 years ago
What does it mean if you titrated the acid to a 'hot pink' color All of these
Murrr4er [49]

Answer: Too much base was added

i guessed

Explanation:

3 0
3 years ago
Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
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