STARTING AMOUNT X How many grams of Na2SO4 are needed to completely precipitate the Ba²+ ions as BaSO4 from 150.0 mL solution of 0.150 M BaCl2 according to the balanced chemical reaction: Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2 NaCl(aq) ADD FACTOR *( ) ANSWER RESET ១ 0.1 6.022 x 1023 208.23 0.01 4.69 1.32 3.20 5.25 0.150 58.44 100 142.04 10 1 2 1000 233.39 150.0 mol BaCl2 mL BaCl2 mL Na2SO4 L solution g Na2SO4 L Na₂SO g BaCl, L BaCl₂ mol Na2SO4

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Chapter1: Chemical Foundations
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STARTING AMOUNT
X
How many grams of Na2SO4 are needed to completely precipitate the Ba²+ ions as
BaSO4 from 150.0 mL solution of 0.150 M BaCl2 according to the balanced
chemical reaction:
Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2 NaCl(aq)
ADD FACTOR
*( )
ANSWER
RESET
១
0.1
6.022 x 1023 208.23
0.01
4.69
1.32
3.20
5.25
0.150
58.44
100
142.04
10
1
2
1000
233.39
150.0
mol BaCl2 mL BaCl2 mL Na2SO4 L solution g Na2SO4
L Na₂SO
g BaCl,
L BaCl₂
mol Na2SO4
Transcribed Image Text:STARTING AMOUNT X How many grams of Na2SO4 are needed to completely precipitate the Ba²+ ions as BaSO4 from 150.0 mL solution of 0.150 M BaCl2 according to the balanced chemical reaction: Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2 NaCl(aq) ADD FACTOR *( ) ANSWER RESET ១ 0.1 6.022 x 1023 208.23 0.01 4.69 1.32 3.20 5.25 0.150 58.44 100 142.04 10 1 2 1000 233.39 150.0 mol BaCl2 mL BaCl2 mL Na2SO4 L solution g Na2SO4 L Na₂SO g BaCl, L BaCl₂ mol Na2SO4
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