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
Newton's first law of motion states that an object at rest will remain at rest unless an unbalanced force acts on it. If you apply balanced forces on the object there would be no net force. The body does not accelerate but instead stays at rest.
Another way to look at this problem is to use Newton's second law of motion. The first law states that 
, where 
is the acceleration
is the net force and 
is the mass of the object.
When F is zero, the acceleration of the object is zero. This means that if the object had a velocity of zero before the balanced forces started acting, the velocity will stay at zero after the balanced forces begin to act. If the object was moving at a constant velocity before the balanced forces started acting on it, it would continue at that constant velocity after the balanced forces begin to act.
The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Using the equation,
Δ
= i
m
where:
Δ = change in freezing point (unknown)
i = Van't Hoff factor
= freezing point depression constant
m = molal concentration of the solution
Molality is expressed as the number of moles of the solute per kilogram of the solvent.
Molal concentration is as follows;
MM KCl = 74.55 g/mol
molal concentration = 
molal concentration = 0.1219m
Now, putting in the values to the equtaion Δ = im we get,
Δ = 2 × 1.86 × 0.1219
Δ = 0.4536°C
So, Δ of solution is,
Δ
= 0.00°C - 0.45°C
Δ = - 0.45°C
Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Learn more about freezing point here;
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<span>The unit cell of amazonite is made of three pairs of faces, each pair different from the others. The faces are oriented at angles to one another that are not 90 degrees. The type of crystal this is is triclinic.
Triclinic crystal system is one of the seven crystal systems, with three basis vectors. </span>
Answer:
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hope this helps!
Explanation:
