Answers
• Volume of the marshmallow:
V = 2.75 in^3 (but, 1 in^3 = 16.39 cm^3)
V = 2.75 × 16.39 cm^3
V = 2.75 × 16.39 cm^3
V = 45.0725 cm^3
• Density:
d = 0.242 g/cm^3
• Mass:
m = d × V
m = (0.242 g/cm^3) × (45.0725 cm^3)
m = (0.242 g/cm^3) × (45.0725 cm^3)
m = 10.907545 g
m ≈ 10.9 g <——— this is the answer.
I hope this helps. =)
Answer:
9.92g
Explanation:
2.50 in31×16.39 cm31 in3×0.242 gcm3=9.92 g
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Calculate the molar solubility of Cd(OH)2 when buffered at a pH = 12.30. The Ksp for Cd(OH)2 is 2.5 x 10-14. Calculate the molar solubility of Cd(OH)2 when buffered at a pH = 12.30. The Ksp for Cd(OH)2 is a.2.5 x 10-14 M. b. 8.5 x 10-6 M c. 6.3 x 10-11 M d. 1.3 x 10-12 M e. 5.0 x 10-2 M f. 1.8 x 10-5 M
Help me as soon as possible I’m gonna dieeee
Answers
♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫
➷ 1) Enzyme A
[This is because a pH below 7 is acidic and the optimum pH is 1 for this enzyme]
2) Enzyme C
[pH 7 is neutral]
3) Enzyme D
[A pH above 7 is 'basic']
4) Enzyme B
[It is still below 7 but it is closer to neutral so it is just 'acidic']
✽
➶ Hope This Helps You!
➶ Good Luck (:
➶ Have A Great Day ^-^
↬ ʜᴀɴɴᴀʜ ♡
1st one is A
2nd one is C
3rd one is D
4th one is B
(Brainliest plz)
Answers
Answers
Answer:
solubility of X in water at 17.0 is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0 can be calculated using the information given.
Let's assume solubility of X in water at 17.0 is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y = = 0.11
Hence solubility of X in water at 17.0 is 0.11 g/mL.
Answers
Answer:
35.6 g of W, is the theoretical yield
Explanation:
This is the reaction
WO₃ + 3H₂ → 3H₂O + W
Let's determine the limiting reactant:
Mass / molar mass = moles
45 g / 231.84 g/mol = 0.194 moles
1.50 g / 2 g/mol = 0.75 moles
Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.
Let's make rules of three:
1 mol of tungsten(VI) oxide needs 3 moles of H₂
Then 0.194 moles of tungsten(VI) oxide would need (0.194 .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)
3 moles of H₂ need 1 mol of WO₃ to react
0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles
It's ok. I do not have enough WO₃.
Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.
Let's convert the moles to mass (molar mass . mol)
0.194 mol . 183.84 g/mol = 35.6 g
Answers
Answer:
Polarity
Cohesion
Adhesion
High Specific Heat
Explanation:
Answers
Explanation:
CH2)3(g)CH3CH=CH2(g) [(CH2)3], M time, min
0.128 0
6.40×10-2 14.4
3.20×10-2 28.8
1.60×10-2 43.2
(1) What is the half-life for the reaction starting at t=0 min? min
Half life is the amount of time required for a substance to decay by half of it's initial concentration.
Starting form 0, the initial concentration = 0.128
After 14.4 mins, the final concentration is now exactly half of the initial concentration. This means 14.4 min is the half life starting from t=0min
What is the half-life for the reaction starting at t=14.4 min?
Starting form 14.4min, the initial concentration = 6.40×10-2
After 14.4 mins (28.8 - 14.4), the final concentration is now exactly half of the initial concentration. This means 14.4 min is the half life starting from t=14.4min
Does the half-life increase, decrease or remain constant as the reaction proceeds?
The half life is a constant factor, hence it remains constant as the reaction proceeds.
(2) Is the reaction zero, first, or second order?
Because the half life is independent of the concentration, it is a first order reaction.
In a zero order reaction, the half life Decreases as the reaction progresses; as concentration decreases.
In a first order reaction, the half life Increases with decreasing concentration.
(3) Based on these data, what is the rate constant for the reaction? min-1
The realtionship between the half life and rate onstant is;
k = 0.693 / half life
k = 0.693 / 14.4
k = 0.048125 min-1