Answer:
Allows diffusion of sub...(true)
its found in both anim...(true)
produces en...(false)
Answer:
potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol
to get 1000 ml
Molar concentration = Mass concentration/Molar Mass
mass concentration = molar concentration x molar mass
mass concentration=0.1 M,
molar mass= 204.233 g/mol
so to get 1L
mass conc = 204.233 x 0.1
= 20.4233g for 1L or 1000 ml
to get 6.00 ml
if 20.4233g is for 1000ml
then to 6.00 ml
= 20.4233 x 6 / 1000
= 0.123g for 6.00 ml
according to the equation below
NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)
number of moles of NaOH is equal to that of KHP
so the same amount will be needed too, which is
= 0.123g
A rock is any naturally occurring solid mass or aggregate of minerals or mineraloid matter. It is categorized by the minerals included, its chemical composition and the way in which it is formed
Answer: (E) 300 bq
Explanation:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.
Half life is represented by 
Half life of Thallium-208 = 3.053 min
Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. 
, after second half life, the activity would be reduced to half of 1200 i.e. 
, and after third half life, the activity would be reduced to half of 600 i.e. 
,
Thus the activity 9 minutes later is 300 bq.
Answer:
The answer is
<h3>2.53 × 10²⁴ molecules</h3>
Explanation:
The number of molecules present can be found by using the formula
<h3>N = n × L</h3>
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 4.21 × 6.02 × 10²³
We have the final answer as
<h3>2.53 × 10²⁴ molecules</h3>
Hope this helps you
