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
Answer:
The answer is E. All of the statements describe the anomeric carbon.
Explanation:
When a sugar switches from its open form to its ring form, the carbon from the carbonyl (aldehyde if it is an aldose, or a ketone in the case of a ketose) suffers a nucleophilic addition by one of the hydroxyls in the chain, preferably one that will form a 5 or 6 membered ring after the reaction.
As such, the anomeric carbon will have two oxygens attached (The original one and the one that bonded when the ring closed).
It will be chiral, given that it has 4 different groups attached. (-OR,-OH,-H and -R, where R is the carbon chain).
The hydroxyl group can be in any position (Above of below the ring), depending on with side the addition took place. (See attachment)
It is the carbon of the carbonyl in the open-chain form of the sugar, because it is the only one that can react with the Hydroxyls.
Answer:
B Genes determine specific traits while the chromosomes contain these genes
P₄0₆
+ 20₂
⇒ P₄0₁₀
Explanation:
The overall equation for the reaction that produces P₄0₁₀ is :
P₄0₆ + 20₂ ⇒ P₄0₁₀
Now let us derive this equation:
Given equations:
P₄ + 30₂ ⇒ P₄0₆ equation 1;
P₄ + 50₂ ⇒ P₄0₁₀ equation 2;
To get the overall combined equation, the equation 1 must be reversed and added to equation 2:
P₄0₆ ⇒ P₄ + 30₂ equation 3
+
equation 2:
P₄ + 50₂ + P₄0₆ ⇒ P₄0₁₀ + P₄ + 30₂
cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;
P₄0₆ + 20₂ ⇒ P₄0₁₀
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Answer:
After 2.0 minutes the concentration of N2O is 0.3325 M
Explanation:
Step 1: Data given
rate = k[N2O]
initial concentration of N2O of 0.50 M
k = 3.4 * 10^-3/s
Step 2: The balanced equation
2N2O(g) → 2 N2(g) + O2(g)
Step 3: Calculate the concentration of N2O after 2.0 minutes
We use the rate law to derive a time dependent equation.
-d[N2O]/dt = k[N2O]
ln[N2O] = -kt + ln[N2O]i
⇒ with k = 3.4 *10^-3 /s
⇒ with t = 2.0 minutes = 120s
⇒ with [N2O]i = initial conc of N2O = 0.50 M
ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)
ln[N2O] = -1.101
e^(ln[N2O]) = e^(-1.1011)
[N2O} = 0.3325 M
After 2.0 minutes the concentration of N2O is 0.3325 M
