2(CH3)2O3 + 2H2O ---> 4 CH3COOH is the balanced equation.
Answer:
Motile bacteria have flagella, while nonmotile bacteria do not.
Explanation:
This is what I got. Hope it helps :)
2 moles of sodium hydroxide will be needed.
<h3><u>Explanation</u>:</h3>
Sodium hydroxide is a compound which is a base and nitric acid is the acid. The formula of the nitric acid is HNO3 and that of sodium hydroxide is NaOH.
The reaction between them are
NaOH +HNO3 =NaNO3 +H2O.
So here we can see that 1 mole of sodium hydroxide reacts with 1 mole of nitric acid to produce 1 mole of sodium nitrate and 1 mole of water.
So for 2 moles of nitric acid, 2 moles of sodium hydroxide will be required.
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>
