Cation is formed when an atom of mercury (Hg) loses two electrons.
Answer:
The answer to your question is 3 ml
Explanation:
Data
Dosage = 9.0 mg/ kg
Child's weight = 42.9 pounds
Suspension = 60 mg/ml
milliliters = ?
Process
1.- Convert the weight to kg
1 pound ------------------- 0.453 kg
42.9 pounds --------------- x
x = (42.9 x 0.453) / 1
x = 19.43 kg
2.- Calculate the milligrams the child needs
1 kg of weight ------------ 9 mg
19.43 kg ---------------------- x
x = (19.43 x 9) / 1
x = 174.87 mg of oxcarbazepine
3.- Calculate the milliliters needed
60 mg of suspension ------------- 1 milliliters
174.87 mg -------------- x
x = (174.87 x 1) / 60
x = 2.9 ml ≈ 3 ml
Answer:
Q14: 17,140 g = 17.14 kg.
Q16: 504 J.
Explanation:
<u><em>Q14:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).
m is the mass of the ice (m = ??? g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).
∵ Q = m.c.ΔT
∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)
∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.
<u><em>Q16:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 12.0 g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).
∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.
Answer:
B. Four moles of water were produced from this reaction.
Explanation:
I took the test and got it correct
O2 gas, where there are two Oxygen atoms which are covalently bonded together
