Answer:
c) Check to see if there's a better medication for the patient's problem.
d) Dispense an extra dose to save the patient from having to return in case of loss or damage to one of the doses.
e ) Compare the label on the medication with the order from the physician .
Answer:
Final T = 64.0°C.
Explanation:
- The amount of heat absorbed by Na (Q) can be calculated from the relation:
<em>Q = m.c.ΔT.</em>
where, Q is the amount of heat absorbed by Na (Q = 1840 J),
m is the mass of Na (m = 68.0 g),
c is the specific heat capacity of Na (c = 1.23 J/g °C),
ΔT is the temperature difference (final T - initial T) (ΔT = final T - 42.0°C).
∵ Q = m.c.ΔT.
∴ (1840 J) = (68.0 g)(1.23 J/g °C)(final T - 42.0°C)
(final T - 42.0°C) = (1840 J)/(68.0 g)(1.23 J/g °C) = 22.0°C.
<em>∴ final T</em> = 22.0°C + 42.0°C = <em>64.0°C.</em>
I am not sure but rocky bodies ranging from large to small in size
It is practical research of Boyle’s law..
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Answer:
(FeSCN⁺²) = 0.11 mM
Explanation:
Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2
M (Fe(NO₃)₃ = 0.200 M
V (Fe(NO₃)₃ = 10.63 mL
n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol
M (KSCN) = 0.00200 M
V (KSCN) = 1.42 mL
n (KSCN) = 0.00200 * 1.42 = 0.00284 mmol
Total volume = V (Fe(NO₃)₃ + V (KSCN)
= 10.63 + 1.42
= 12.05 mL
Limiting reactant = KSCN
So,
FeSCN⁺² = 0.00284 mmol
M (FeSCN⁺²) = 0.00284/12.05
= 0.000236 M
Excess reactant = (Fe(NO₃)₃
n(Fe(NO₃)₃ = 2.126 mmol - 0.00284 mmol
=2.123 mmol
For standard 2:
n (FeSCN⁺²) = 0.000236 * 4.63
=0.00109
V(standard 2) = 4.63 + 5.17
= 9.8 mL
M (FeSCN⁺²) = 0.00109/9.8
= 0.000111 M = 0.11 mM
Therefore, (FeSCN⁺²) = 0.11 mM
