Answer:
5.59 %
Explanation:
From the question given above, the following data were obtained:
Observed value of density = 2.85 g/cm³
True value of density = 2.699 g/cm³
Percentage error =.?
The percentage error of the student can be obtained as follow:
Percentage error = |Observed value – True value|/True value × 100
Percentage error = |2.85 – 2.699|/2.699
Percentage error = 0.151/2.699 × 100
Percentage error = 5.59 %
Therefore, the percentage error of the student is 5.59 %.
Answer:
After 18 hours, the amount of pure technetium that will be remaining is 12.5 grams
Explanation:
To solve the question, we note that the equation for half life is as follows;
Where:
N(t) = Quantity of the remaining substance = Required quantity
N₀ = Initial radioactive substance quantity = 100 g
t = Time duration = 18 hours
= Half life of the radioactive substance = 6 hours
Therefore, plugging in the values, we have;
Therefore, after 18 hours, the amount of pure technetium that will be remaining = 12.5 grams.
Answer:
32.5g of sodium carbonate
Explanation:
Reaction of sodium carbonate (Na₂CO₃) with Mg²⁺ and Ca²⁺ as follows:
Na₂CO₃(aq) + Ca²⁺(aq) → CaCO₃(s)
Na₂CO₃(aq) + Mg²⁺(aq) → MgCO₃(s)
<em>1 mole of carbonate reacts per mole of the cations.</em>
<em />
To know the mass of sodium carbonate we must know the moles of carbonate we need to add based on the moles of the cations:
<em>Moles Mg²⁺:</em>
2.91L * (0.0661 moles MgCl₂ / 1L) = 0.192 moles MgCl₂ = Moles Mg²⁺
<em>Moles Ca²⁺:</em>
2.91L * (0.0396mol Ca(NO₃)₂ / 1L) = 0.115 moles Ca(NO₃)₂ = Moles Ca²⁺
That means moles of sodium carbonate you must add are:
0.192 moles + 0.115 moles = 0.307 moles sodium carbonate.
In grams (Using molar mass Na₂CO₃ = 105.99g/mol):
0.307 moles Na₂CO₃ * (105.99g / mol) =
<h3>32.5g of sodium carbonate</h3>
<u>Answer: </u>The correct answer is Option B.
<u>Explanation:</u>
The given elements that is sodium and sulfur are the elements which belong to the same period that is third period.
As, the electrons are added up to the same shell across a period, the number of protons also increase and hence, they attract the electrons more effectively. This means that the effective nuclear charge also increases and the size of the sulfur thus shrink and hence, the size (atomic radius) of sodium is the largest amongst third period.
Therefore, the correct answer is Option B.
