What is the oxidation state of an individual bromine atom in NaBrO3? Part A What

Question

What is the oxidation state of an individual bromine atom in NaBrO3?

Part A

What mass of silver chloride can be produced from 1.68 L of a 0.243 M solution of silver nitrate?

Express your answer with the appropriate units.

Part B

The reaction described in Part A required 3.43 L of magnesium chloride. What is the concentration of this magnesium chloride solution?

Express your answer with the appropriate units

Part A

Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 2.01×103 kg of sulfuric acid solution?

Express your answer with the appropriate units

Part A

Which of the following substances could the unknown be: KOH, NH3, HNO3, KClO2, H3PO3, CH3COCH3(acetone)?

Check all that apply.

Part A

Which element is oxidized in this reaction?

Fe2O3+3CO2Fe+3CO2

Enter the chemical symbol of the element

Part C

Which element is reduced in this reaction?

2Cr(OH)3+3OCl+4OH2CrO42+3Cl+5H2O

Enter the chemical symbol of the element

Explanation / Answer

Na=+1(1st grop elements have +1)
O=-2(mostly -2 except peroxides)
O3=3*(-2)=-6
let oxidation state of Br is X
then+1+X-6=0
X-5=0
X=5

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2AgNO3(aq)+MgCl2(aq)=2AgCl(s)+Mg(NO3)2

Part-A

amount of AgNO3 = 0.243 x 1.68 = 0.4082mol

amount of AgCl precipitated = 0.4082mol

molar mass of AgCl = 107.86 + 35.453 = 143.313g/mol

Mass of AgCl precipitated = 0.4082 x 143.313 = 58.50g

Part-B

amount of MgCl2 = n(AgCl)/2 = 0.2040mol

this is dissolved in 3.43L so the concentration is

c(MgCl2) = n(MgCl2)/V = 0.2040/3.43 = 0.059mol/L

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Part-A

You have 2010kg of H2SO4 at
Balanced equation:
H2SO4 + Na2CO3 Na2SO4 + CO2 + H2O
1mol H2SO4 reacts with 1mol Na2CO3
Molar mass H2SO4 = 98.08 g/mol

Therefore you have 2010*1000 / 98.08 = 20,493.47moles of H2SO4

You need the same moles of Na2CO3 = 20,493.47moles

Molar mass Na2CO3 = 105.9884 g/mol (anhydrous)124.00 g/mol (monohydrate) and 286.14 g/mol (decahydrate)

For simplicity we will work with the anhydrous product:
20,493.47 * 105.9884/1000 = 2172kg Na2CO3 anhydrous required.

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Part-A

KOH- base
NH3- weak base
KCLO2- salt
H3PO3- weak acid
Acetone- organic compound

The only one it could be is H3PO3, because its the only acid there.

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Fe2O3+3CO2Fe+3CO2

Carbon is oxidized

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Part-C

2Cr(OH)3+3OCl+4OH2CrO42+3Cl+5H2O

Cl is reduced

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