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>
Making repeated separations of the various substances in the pitchblende, Marie and Pierre used the Curie electrometer to identify the most radioactive fractions. They thus discovered that two fractions, one containing mostly bismuth and the other containing mostly barium, were strongly radioactive.
<h3>What was surprising about pitchblende?</h3>
Since it was no longer appropriate to call them “uranic rays,” Marie proposed a new name: “radioactivity.”
Even more surprising, Marie next found that a uranium ore called pitchblende contained two powerfully radioactive new elements: polonium, which she named for her native Poland, and radium.
<h3>Why is radium more radioactive than uranium?</h3>
It is 2.7 million times more radioactive than the same molar amount of natural uranium (mostly uranium-238), due to its proportionally shorter half-life.
Learn more about highly radioactive elements here:
<h3>
brainly.com/question/10257016</h3><h3 /><h3>#SPJ4</h3>
Here is the complete question.
Benzalkonium Chloride Solution ------------> 250ml
Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.
Sig: Dilute 10ml to a liter and apply to affected area twice daily
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL
Answer:
(B) 29.4 mL
Explanation:
1 L = 1000 mL
1:200 solution implies the 
in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be 
200mL × 1g = 1000 mL × x(g)
x(g) = 
x(g) = 0.2 g
That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.
∴ 
y(g) = 
y(g) = 5g of benzalkonium chloride.
Now, at 17% concentrate contains 17g/100ml:
∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;
= 
z(mL) = 
z(mL) = 29.41176 mL
≅ 29.4 mL
Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride
