When a change in PH = 10^-ΔPH
so the change = 10^-3.2
change depends on two factor 0.00063 (10^-3.2)and factor 1585 (10^3.2)depending on the way which the change goes.if PH change from PH=0 to PH= 3.2 so the change is decreasing from concentration from 1 mol to 0.00063 and if PH change from PH = 3.2 to PH=0 so the change is increasing by a factor of 1585.
Answer:
(a) r = 6.26 * 10⁻⁷cm
(b) r₂ = 6.05 * 10⁻⁷cm
Explanation:
Using the sedimentation coefficient formula;
s = M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle
s = M ( 1 - Vρ) / N*6πnr
making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs
Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s
r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)
r = 6.26 * 10⁻⁷cm
b. Using the formula r₂/r₁ = s₁/s₂
s₂ = 0.035 + 1s₁ = 1.035s₁
making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁
r₂ = 6.3 * 10⁻⁷cm / 1.035
r₂ = 6.05 * 10⁻⁷cm
It’s A. Land heats up and cools quickly then water
Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g
<h3>Avogadro's hypothesis </h3>
6.02×10²³ atoms = 1 mole of arsenic
But
1 mole of arsenic = 75 g
Thus, we can say that:
6.02×10²³ atoms = 75 g of arsenic
<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>
6.02×10²³ atoms = 75 g of arsenic
Therefore,
1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)
1.23×10²⁰ atoms = 0.0153 g of arsenic
Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic
Learn more about Avogadro's number:
brainly.com/question/26141731
Answer:
Explanation:
In this case, we can start with the <u>formula of Platinum (II) Chloride</u>. The cation is the atom at the left of the name (in this case 
) and the anion is the atom at the right of the name (in this case 
). With this in mind, the <u>formula would be</u> 
.
Now, if we used <u>metallic copper</u> we have to put in the reaction only the <u>copper atom symbol</u> 
. So, we have as reagents:
The question now is: <u>What would be the products?</u> To answer this, we have to remember <u>"single displacement reactions"</u>. With a general reaction:
With this in mind, the reaction would be:
I hope it helps!
