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NARA [144]
4 years ago
8

Did barium chloride and barium sulfate produce similar colored flames? yes, both compounds produced pale yellow-green flames. no

, barium chloride produced pale yellow-green flames and barium sulfate produced red flames. no, barium chloride produced purple flames and barium sulfate produced yellow flames. yes, both compounds produced red flames.
Chemistry
2 answers:
Llana [10]4 years ago
7 0
Answer = A

Explanation:
                     
When Ba⁺² is subjected to flame it burns yellow-green under the flame. This colour is generated due to the fact that on exposure to heat the electrons of metal absorbs energy and gets excited to higher energy level, and bounce back to ground state with the elimination of energy absorbed. This energy of particular wavelength falls in visible region and gives yellow-green colour. When compounds of Barium chloride and Barium Sulphate are subjected to flame the give same colour. (Reference for BaSO₄ = https://carnicominstitute.org/wp/barium-tests-are-positive/)
Rudiy274 years ago
5 0

The answer is calcium chloride

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Part IV. Limiting Reactants! A Challenge Problem!
Alexxandr [17]

Answer:

a. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

b. Fe2O3 is the limiting reactant.

c. 6.30 grams Fe

d. 52.6 %

Explanation:

Step 1: Data given

Mass of iron(III) oxide Fe2O3 = 9.00 grams

Mass of aluminium = 4.00 grams

Molar mass Fe2O3 = 159.69 g/mol

Aluminium molar mass = 26.98 g/mol

Step 2: The balanced equation

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

Step 3; Calculate Moles

Moles = mass / molar mass

Moles Fe2O3 = 9.00 grams / 159.69 g/mol

Moles Fe2O3 = 0.0564 moles

Moles Al = 4.00 grams / 26.98 g/mol

Moles Al = 0.148 moles

Step 4: Calculate limiting reactant

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

Fe2O3 is the limiting reactant. It will completely be consumed (0.0564 moles).  Al is in excess. There will react 0.0564*2 = 0.1128 moles

There will remain 0.148 - 0.1128 = 0.0352 moles Al

Step 5: Calculate moles Fe

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

For 0.0564 moles Fe2O3 we'll have 2*0.0564 = 0.1128 moles Fe

Step 6: Mass of Fe

Mass Fe = 0.1128 moles * 55.845 g/mol

Mass Fe = 6.30 grams

Step 7: If you carried out this reaction and it actually produced 0.475 mL of molten iron (r = 6.98 g/mL), what is the percent yield of this reaction?

Density = mass / volume

Mass = density * volume

Mass = 6.98 g/mL * 0.475 mL

Mass = 3.3155 grams

Percent yield = (actual mass / theoretical mass) * 100%

Percent yield = (3.3155 /6.30 ) * 100 %

Percent yield = 52.6 %

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Explanation:

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