Given:
Stock dose/concentration of 20% Acetylcysteine (200 mg/mL)
150 mg/kg dose of Acetylcysteine
Weight of the dog is 13.2 lb
First we must convert 13.2 lb to kg:
13.2 lb/(2.2kg/lb) = 6 kg
Then we must calculate the dose:
(150 mg/kg)(6kg) = 900 mg
Lastly, we must calculate the dose in liquid form to be administered:
(900 mg)/(200 mg/mL) = 4.5 mL
Therefore, 4.5 mL of 20% Acetylcysteine should be given.
An altered state of consciousness can be defined as any state of consciousness that deviates from normal waking consciousness, in terms of marked differences in our level of awareness, perceptions, memories, thinking, emotions, behaviours and sense of time, place and self-control.
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<span>Van der waal or ideal eqn is given by PV = NRT; P = NRT/ V.
Where N = 1.335 is the number of moles. T = 272K is temperature. V = 4.920L is the volume. And R = 0.08205L. Substiting the values into the eqn; we have,
P = (1.331* 0.08205 * 272)/ 4.920 = 29.7047/ 4.920 = 6.03atm.</span>
The answer is C) <span>Helium forms the solar core, which continually increases in size.
When the hydrogen is </span><span>produced by the conversion and made into Helium the core contains the helium as well as producing it. Which in turn continually increases the size of the solar core.</span>
Answer:
a. 0.80V b. 2Ag⁺(aq) + H2(g) ⇄ 2Ag(s) +2H⁺(aq) c. i) 0.88 ii) 1.03 d. Cell is a ph meter with the potential being a function of hydrogen ion concentration
Explanation:
a. The two half cell reactions are
1. 2H⁺(aq) +2e⁻ → H₂(g) Eanode = 0.00V
2. Ag⁺(aq) + e⁻ → Ag(s) Ecathode = 0.80V
The balanced cell reaction is
2Ag(aq)⁺ + H₂(g) ⇄ 2Ag(s) + 2H⁺(aq)
therefore Ecell = Ecathode - Eanode = 0.80 - 0.00 = +0.80V
b. Since the Ecell is positive, the spontaneous cell reaction under standar state conditions is
2Ag(aq)⁺ + H₂(g) ⇄ 2Ag(s) + 2H⁺(aq)
c. Use Nernst Equation
E = Ecell - (0.0592/n)log([H⁺]/[Ag⁺]²[P H₂]), where n is the number of moles of Ag and P H₂= 1.0 atm
i) E = 0.80 - (0.0592/2)log(4.2x10^-2)/(1.0)²(1.0) = 0.88V
ii) E = 0.80 - (0.0592/2)log(9.6x10^-5)/(1.0)²(1.0) = 1.03V
d . From the above calculation we can conclude that the cell acts as a pH meter as a change in hydrogen ion concentration results in a change in the potential of the cell. A change of ph of 2.64 changes the E of cell by 0.15 V.
