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3 years ago

Write the chemical formula for sodium hypobromite.

Chemistry

1 answer:

irga5000 [103]3 years ago

6 0

Answer:

NaOBr (or) Na⁺ ⁻OBr

Explanation:

The Oxo-Acids of Bromine are as follow,

Hypobromous Acid = HOBr

Bromous Acid = HOBrO

Bromic Acid = HBrO₃

Perbromic Acid = HBrO₄

When these acids are converted to their conjugate bases their names are as follow,

Hypobromite = ⁻OBr

Bromite = ⁻OBrO

Bromate = ⁻OBrO₂

Perbromate = ⁻OBrO₃

According to rules, the positive part of ionic compound is named first and the negative part is named second. So, Sodium Hypobromite has a chemical formula of Na⁺ ⁻OBr or NaOBr.

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A 55.0g sample of iron (III) filings is reacted with 23.8g of powdered sulfur (S8). How much iron (III) sulfide in moles would b

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Answer:

0.744 mol

Explanation:

the balanced equation for the reaction is

8Fe + S₈ ---> 8FeS

molar ratio of Fe to S₈ is 8:1

number of moles of Fe - 55.0 g / 56 g/mol = 0.98 mol

number of moles of S - 23.8 g / 256 g/mol = 0.093 mol

if we are to assume that S₈ is the limiting reactant

if 1 mol of S₈ reacts with 8 mol of Fe

then 0.093 mol of S₈ reacts with - 8 x 0.093 mol = 0.744 mol of Fe

however there's 0.98 mol of Fe present but only 0.744 mol of Fe is needed

therefore Fe is in excess and S₈ is the limiting reagent

molar ratio of S₈ to FeS is 1:8

then 0.093 mol of S₈ reacts with - 8 x 0.093 = 0.744 mol of FeS

number of FeS moles produced is 0.744 mol

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What is the formula for the covalent compounds sulfur pentaoxide and hydrogen monobromide?

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3 years ago

A sample of 9.27 g9.27 g of solid calcium hydroxide is added to 38.5 mL38.5 mL of 0.500 M0.500 M aqueous hydrochloric acid. Writ

navik [9.2K]

Answer: The excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

Explanation:

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

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

For calcium hydroxide:

Given mass of calcium hydroxide = 9.27 g

Molar mass of calcium hydroxide = 74.093 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium hydroxide}=\frac{9.27g}{74.093g/mol}=0.125mol$

To calculate the moles of a solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Volume of hydrochloric acid = 38.5mL = 0.0385 L (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.500 moles/ L

Putting values in above equation, we get:

$0.500mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0385L}\\\\\text{Moles of hydrochloric acid}=0.01925mol$

For the given chemical equation:

$2HCl(aq.)+Ca(OH)_2(s)\rightarrow CaCl_2(s)+2H_2O(l)$

Here, the solid salt is calcium chloride.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.01925 moles of hydrochloric acid will react with = $\frac{1}{2}\times 0.01925=0.009625moles$ of calcium hydroxide.

As, given amount of calcium hydroxide is more than the required amount. So, it is considered as an excess reagent.

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

Amount of excess reagent (calcium hydroxide) left = 0.125 - 0.01925 = 0.115375 moles

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.01925 moles of hydrochloric acid will produce = $\frac{1}{2}\times 0.01925=0.009625moles$ of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 110.98 g/mol

Moles of calcium chloride = 0.009625 moles

Putting values in equation 1, we get:

$0.009625mol=\frac{\text{Mass of calcium chloride}}{110.98g/mol}\\\\\text{Mass of calcium chloride}=1.068g$

Hence, the excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

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3 years ago

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