“Star cluster” is a generic way for astronomers to refer to a group of stars that formed from the same material and are gravitationally bound for at least some period of time. There are two major types of star clusters — globular clusters and open clusters — and they are actually quite different.
The total pressure of the mixture of gases is
equal to the sum of the pressure of each gas as if it is alone in the
container. The partial pressure of a component of the mixture is said to be
equal to the product of the total pressure and the mole fraction of the
component in the mixture.<span>
Partial pressure of hydrogen gas = 1.24 atm x
.25 = 0.31 atm
<span>Partial pressure of the remaining = 1.24 atm x
(1-.25) = 0.93 atm </span></span>
She said that the exposure to the radiation in the microwave would have cooked the animal's internal organs. She said: 'It is a horrific case in the fact that the death of the cat would have been prolonged and it is unimaginable what it would have gone through taking some time to die.
Answer:
pH = 5.35
Explanation:
Given 1.60 grams sodium acetate (NaOAc(aq))*** added to 50ml of 0.10M acetic acid (HOAc(aq)) solution.
Applying common ion effect keeping in mind that the addition of NaOAc provides the common-ion (OAc⁻).
HOAc(aq) ⇄ H⁺(aq) + OAc⁻(aq)
I 0.10m 1.32 x 10⁻³M ≈ ∅M* (1.6g/82.03g/mol) / 0.050L = 0.39M
C -x +x 0.39M + x ≈ 0.39M**
E 0.10M - x x 0.39M
≈ 0.10M
Ka = [H⁺][OAC⁻]/[HOAC] => [H⁺] = Ka·[HOAc] / [OAc⁻]
[H⁺] = (1.75 X 10⁻⁵)(0.10) / (0.39) = 4.5 x 10⁻⁶M
∴ pH = -log[H⁺] = -log(4.5 x 10⁻⁶) = -(-5.35) = 5.35
_______________________________________________
* [H⁺] before adding NaOAc = SqrRt(Ka · [HOAc]) = SqrRt(1.75 x 10⁻⁵· 0.10) = 1.32 x 10⁻³M. Since this concentration value is so small, the initial [H⁺] is assumed to be zero molar (∅M).
** The added [H⁺] is negligible and dropped in the ICE table. That is, adding ~[H⁺] in the order of 10⁻³M does not change the H⁺ ion concentration sufficiently to affect problem outcome and is therefore dropped in the ICE table.
*** Acetic Acid and Sodium Acetate are frequently written HOAc and NaOAc where the OAc⁻ anion is the acetate ion (CH₃COO⁻) for brevity.
Answer:
The amount of isopropyl alcohol contained in 150 ml of the solution is 117.15 grams
Explanation:
The density of a substance is the mass per unit volume, therefore, we have;
The density of the isopropyl alcohol = 0.785 g/ml at 25°C
The density = mass/volume
Mass = Density × Volume
The mass of the 150 ml of isopropyl alcohol is therefore;
Mass = 0.785 g/ml × 150 ml = 117.15 g
The amount in grams contained in 150 ml of sample = 117.15 g.
