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Rasek [7]
3 years ago
8

An order is given to administer methylprednisolone, an anti-inflammatory drug, by IV at a rate of 39 mg every 30 min. The IV bag

contains 125 mg of methylprednisolone in every 2 mL. What should the flow rate be in milliliters per minute (mL/min)? flow rate=_______ mL/min
Chemistry
1 answer:
Korvikt [17]3 years ago
3 0

Answer:

The Flow rate = 0.0208 mL/min

Explanation:

Data provided:

Rate of dose = 39 mg every 30 min = (39/30) mg/min  = 1.3 mg/min

also,

125mg of methylprednisolone is present in every 2 mL

thus,

concentration = (125/2) mg/mL = 62.5 mg/mL

Now,

The flow rate is given as:

Flow rate = Rate / Concentration

on substituting the respective values, we get

Flow rate = (1.3 mg/min) / (62.5mg/mL)

or

The Flow rate = 0.0208 mL/min

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2 years ago
A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at
lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

<u> Step 2:</u> Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25  = 8.2 °C

q = 1200 * 4.184 * 8.2 =  41170.56 J

<u>Step 4</u>: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J  = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

8 0
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