Answer:
Explanation:
molecular weight of creatinine = 131
1 mg of creatinine = 1 x 10⁻³ / 131 = 7.63 x 10⁻⁶ mole of creatinine.
volume of solution = .1 L
mass of blood solution = .1 x 1025
= 102.5 g
mass of solvent = 102.5 g approximately
= .1025 kg
molality = mole of solute / mass of solvent in kg
= 7.63 x 10⁻⁶ / .1025 kg
= 74.44 x 10⁻⁶ .
Osmotic pressure :---
π V / T = nR π is osmotic pressure , V is volume of solution in liter , T is absolute temperature , n is molality .
π x .1 / 298 = 74.44 x 10⁻⁶ x .082
π = 18.19 x 10⁻³ atm
Answer:
The energy consumed by animals in the form of glucose is conserved because it is transformed into chemical energy as carbon dioxide is produced during respiration.
Explanation:
There's no diagram....but I kinda figured it from the description.
At the anode (A), chloride (Cl-) is oxidized to chlorine. ... At the cathode (C), water is diminished to hydroxide and hydrogen gas. The net procedure is the electrolysis of a fluid arrangement of NaCl into mechanically helpful items sodium hydroxide (NaOH) and chlorine gas.
Answer:
18.2 g.
Explanation:
You need to first figure out how many moles of nitrogen gas and hydrogen (gas) you have. To do this, use the molar masses of nitrogen gas and hydrogen (gas) on the periodic table. You get the following:
0.535 g. N2 and 1.984 g. H2
Then find out which reactant is the limiting one. In this case, it's N2. The amount of ammonia, then, that would be produced is 2 times the amount of moles of N2. This gives you 1.07 mol, approximately. Then multiply this by the molar mass of ammonia to find your answer of 18.2 g.
Answer:
11.12 → pH
Explanation:
This is a titration of a weak base and a strong acid.
In the first step we did not add any acid, so our solution is totally ammonia.
Equation of neutralization is:
NH₃ + HCl → NH₄Cl
Equilibrium for ammonia is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ Kb = 1.8×10⁻⁵
Initially we have 50 mL . 0.10M = 5 mmoles of ammonia
Our molar concentration is 0.1 M
X amount has reacted.
In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.
Expression for Kb is : x² / (0.1 - x) = 1.8×10⁻⁵
As Kb is so small, we can avoid the x to solve a quadratic equation.
1.8×10⁻⁵ = x² / 0.1
1.8×10⁻⁵ . 0.1 = x²
1.8×10⁻⁶ = x²
√1.8×10⁻⁶ = x → 1.34×10⁻³
That's the value for [OH⁻] so:
1×10⁻¹⁴ = [OH⁻] . [H⁺]
1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺] → 7.45×10⁻¹²
- log [H⁺] = pH
- log 7.45×10⁻¹² = 11.12 → pH
