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Setler79 [48]
2 years ago
11

PLEASE HELP WILL GIVE BRAINLIEST

Chemistry
1 answer:
goblinko [34]2 years ago
6 0

Answer:

It traveled a different distance because one has 1 spring the other has 2 spring and the last has3 spring

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A flash contains 85.6g of C7H5NO3S (Saccharine) in 2.00 litre of a solution. What is the molarity?
larisa86 [58]

Answer:

0.234 M

Explanation:

C- 12.009 x 7

H- 1.001 x 5

N- 14.006

O- 16 x 3

S- 32.059

___________+

183.133 g/mol

\frac{1mol}{183.133g} *\frac{85.6g}{2.00L}  = 0.234 M Cancel out the grams mol/L equals molarity. Lowest significant figure is 3

7 0
3 years ago
What are miscible liquids?
sasho [114]

Answer:

ones that can be mixed together

Explanation:

like water or ethanol

7 0
3 years ago
Read 2 more answers
the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29
Vera_Pavlovna [14]

Answer:

ΔG = 16.218 KJ/mol

Explanation:

  • dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate
  • ΔG = ΔG° - RT Ln Q

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴  R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

5 0
3 years ago
Please help :) ill mark you as the brainiest hope you have a good day
Hitman42 [59]

Have a wonderful day :) thanks for the points

5 0
3 years ago
Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. be sure your answer has the correct number
Vera_Pavlovna [14]
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. 1mmol = 10^-3 mol Therefore 4.10*10^-5mmol = 4.10*10^-8mol molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below) But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g Mass is = 9.75*10^-7 grams 1µg = 10^-6 g You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4 (*see below) at this point you could have said: 1µg = 10^-6 g therefore you have a solution of 6.29µg per litre, 155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
3 0
2 years ago
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