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3 years ago

How many moles are represented in 213 grams of sodium bromide , NaBr?

Chemistry

1 answer:

Eduardwww [97]3 years ago

5 0

Answer:

The number of mole represented is 2.07 moles

Explanation:

Using the formula

n = m/Mm

n - mole

m - mass

Mm- molar mass

Since the mass is given to be 213g

So lets calculate the molar mass of sodium bromide (NaBr)

The molar mass of Na = 22.99

Br = 79.904

Molar mass = 22.99+79.904

= 102.894g/mol

Using the formula n = m/Mm

n = 213g/ 102.894g/mol

n = 2.07moles

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Magnesium hydroxide [Mg(OH)2] is an ingredient in some antacids. How many grams of Mg(OH)2 are needed to neutralize the acid in

Allushta [10]

Answer:

1.95g of Mg(OH)2 are needed

Explanation:

Mg(OH)2 reacts with HCl as follows:

Mg(OH)2 + 2 HCl → MgCl2 + 2H2O

Where 1 mole of Mg(OH)2 reacts with 2 moles of HCl

To solve this question we must find the moles of acid. Then, with the chemical equation we can find the moles of Mg(OH)2 and its mass:

Moles HCl:

158mL = 0.158L * (0.106mol / L) = 0.01675 moles HCl

Moles Mg(OH)2:

0.01675 moles HCl * (2mol Mg(OH)2 / 1mol HCl) = 0.3350 moles Mg(OH)2

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0.3350 moles Mg(OH)2 * (58.3197g / mol) =

<h3>1.95g of Mg(OH)2 are needed</h3>

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3 years ago

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nikdorinn [45]

Answer: The isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

Explanation:

Nuclear fission reactions are defined as the reactions in which a heavier nuclei breaks down in two or more smaller nuclei.

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Ba}+^{94}_{36}\textrm{Kr}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 3

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To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 36 + 0

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The isotopic symbol of barium is $_{56}^{139}\textrm{Ba}$

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Sr}+^{143}_{54}\textrm{Xe}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 143 + 3

A = 90

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 54 + 0

Z = 38

The isotopic symbol of strontium is $_{38}^{89}\textrm{Sr}$

Hence, the isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

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Answer:

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