Answer:

Explanation:
Hello,
In this case, we consider oxygen as an ideal gas, for that reason, we use yhe ideal gas equation to compute the moles based on:

Hence, at 3.50 atm and 25 °C for a volume of 2.00 L we compute the moles considering absolute temperature in Kelvins:

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Answer:
B. Cu (s) +Ni(NO3)2 (aq) - 2CuNO3 +Ni (s)
Explanation:
the above reaction is a substitution reaction
If you start with 40.0 grams of the element at noon, 10.0 grams
radioactive element will be left at 2 p.m. The correct answer between
all the choices given is the second choice or letter B. I am hoping that this
answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.
Answer: a. 79.6 s
b. 44.3 s
c. 191 s
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
b) for completion of 32% of reaction
c) for completion of 81 % of reaction
Here we have to calculate the amount of
ion present in the sample.
In the sample solution 0.122g of
ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ + 
(Na)₂SO₄=2Na⁺ + 
Thus, BaCl₂+
= BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of
ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion (
) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of
ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of
ion is present. So in 1 g of BaSO₄
g of
ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of
ion is present.