Lunch of a patient has 3 oz skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk
Energy content of 3 oz skinless chicken is = 110 kcal
Energy content of 3 oz broccoli = 30 kcal
Energy content of 1 medium apple = 60 kcal
Energy content of 1 cup non-fat milk = 90 kcal
So the kilocalories of energy patient obtained from lunch
= 110 kcal+ 30 kcal + 60 kcal + 90 kcal = 290 kcal
Answer:
The solubility of X in water at 17°C is 0.110 g/mL.
Explanation:
The water of a rock pool lined with mineral crystals is a <em>saturated solution</em> of said mineral, this means the concentration of X in those 36 mL is the solubility of compound X in water at 17 °C.
- This means<u> it is possible to calculate said solubility</u>.
The dilution of the sample is not relevant, nor is that 500 mL volume. What's important is that 3.96 g of X form a saturated solution with 36.0 mL of water, so the solubility is:
- 3.96 g / 36.0 mL = 0.110 g/mL
Bond energy is basically how much energy it takes to break that molecule apart. The main things that affect bond energy is electronegativity and the number of bonds. I believe the correct answer is C=O. It <span>has the greatest bond energy, because it is harder to break apart a double bond than a single bond.</span>
The following compounds are soluble in water and will dissociate when dissolved in water.
A) barium hydroxide
B) ethanol
C) glucose
D) silver nitrate
E) dichloromethane
F) postassium chloride
The other compounds ethanol and glucose are also soluble in water but do not dissociate when dissolved in water.
When we have:
Zn(OH)2 → Zn2+ 2OH- with Ksp = 3 x 10 ^-16
and:
Zn2+ + 4OH- → Zn(OH)4 2- with Kf = 2 x 10^15
by mixing those equations together:
Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6
by using ICE table:
Zn(OH)2 + 2OH- → Zn(OH)4 2-
initial 2m 0
change -2X +X
Equ 2-2X X
when we assume that the solubility is X
and when K = [Zn(OH)4 2-] / [OH-]^2
0.6 = X / (2-2X)^2 by solving this equation for X
∴ X = 0.53 m
∴ the solubility of Zn(OH)2 = 0.53 M