Answer:
<h2>mass = 200.23 g</h2>
Explanation:
The density of a substance can be found by using the formula
Since we are finding the mass
<h3>mass = Density × volume</h3>
From the question
Density = 0.81 g/mL
volume = 247.2 mL
Substitute the values into the above formula and solve for the mass
mass = 0.81 × 247.2
= 200.232
We have the final answer as
<h3>mass = 200.23 g to 2 decimal places</h3>
Hope this helps you
Answer:
1.64 moles O₂
Explanation:
Part A:
Remember 1 mole of particles = 6.02 x 10²³ particles
So, the question becomes, how many '6.02 x 10²³'s are there in 9.88 x 10²³ molecules of O₂?
This implies a division of given number of particles by 6.02 x 10²³ particles/mole.
∴moles O₂ = 9.88 x 10²³ molecules O₂ / 6.02 x 10²³ molecules O₂ · mole⁻¹ = 1.64 mole O₂
_______________
Part B needs an equation (usually a combustion of a hydrocarbon).
Answer:
90.35 × 10²³ atoms
Solution:
1 molecules of H20 contains 3 atoms,.
And we know that one mole of any molecule contains 6.023 × 10²³ atoms from Avogadro's number,
hence 5 moles of H20 will contain = 5× 6.023 × 10²³ × 3 atoms = 90.35 × 10²³ atoms!
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C. released or absorbed.
When the energy is released the reaction is called exotermic reaction
When the energy is absorbed, the reaction is called endothermic reaction
An example of exotermic reaction is between water and H2SO4
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
