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zaharov [31]
2 years ago
11

BRAINLIEST! Please help me answer the following questions:

Chemistry
1 answer:
Vlada [557]2 years ago
6 0

Explanation:

1) from the periodic table for example if u looking for K it is from the group 1A so the valancey = +1 that is mean it loses one electron to reach a stable state. and this applied for the 1A , 2A, 3A groups.

if u looking for N it is from group 5A so its valency is -3

so it gain 3 electrons to reach a stable state to complete its orbital to 8 electrons. this applied to 5A, 6A,7A

for the transition metals and simimetals other rules.

2) neutral if the element does not hold any charge like

H2, O2, N2, Fe, Au, Cu ...

isotope : if the 2 elements have same symbol and atomic number but they differ in the mass number because the have different numbers of neutrons.

if the element is charged then it is considered as an ion ex: Fe2+ then it is iron cation ( +ion)

F- then it is fluoride anion (-ion)

if u have any doubt just writ a comment ( ;

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A 22.0 mLmL sample of a 1.16 MM potassium sulfate solution is mixed with 14.8 mLmL of a 0.860 MM barium nitrate solution and thi
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Answer:

The limiting reactant is Ba(NO3)2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

Explanation:

Step 1: Data given

Volume of a 1.16 M potassium sulfate solution (K2SO4) = 22.0 mL = 0.022 L

Volume of a 0.860 M barium nitrate solution (Ba(NO3)2 = 14.8 mL = 0.0148 L

The solid BaSO4 is collected, dried, and found to have a mass of 2.57 grams

Step 2: The balanced equation

K2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2KNO3(aq)

Step 3: Calculate moles

Moles = volume * molarity

Moles K2SO4 = 0.022 L * 1.16 M

Moles K2SO4 = 0.02552 moles

Moles Ba(NO3)2 = 0.0148 L * 0.860 M

Moles Ba(NO3)2 = 0.012728 moles

Step 4: Calculate the limiting reactant

For 1 mol K2SO4 we need 1 mol Ba(NO3)2 to produce 1 mol BaSO4 and 2 moles KNO3

Ba(NO3)2 is the limiting reactant. It will completely be consumed. (0.012728 moles) . K2SO4 is in excess. There will remain 0.02552 - 0.012728 = 0.012792 moles

Step 5: Calculate moles BaSO4

‬For 1 mol K2SO4 we need 1 mol Ba(NO3)2 to produce 1 mol BaSO4 and 2 moles KNO3

For 0.012728 moles Ba(NO3)2 we'll have 0.012728 moles BaSO4

Step 6: Calculate mass BaSO4

Mass BasO4 = moles BaSO4 * molar mass BaSO4

Mass BaSO4 =  0.012728 moles *  233.38 g/mol

Mass BaSO4 = 2.97 grams

Step 7: Calculate the percent yield

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

% yield = ( 2.57 grams / 2.97 grams ) * 100 %

% yield = 86.5 %

The limiting reactant is Ba(NO3)2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

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