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valentina_108 [34]
3 years ago
13

A patient receives 3.3 L of glucose solution intravenously (IV). If 100. mL of the solution

Chemistry
1 answer:
artcher [175]3 years ago
6 0

Answer:

660kcal

Explanation:

The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.  

We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:  

3.3L * (1000mL / L) * (5g/100mL)= 165 g.

If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.

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117.22 g are needed to react with an excess of Fe2O3 to produce 156.2 g of Fe.

Explanation:

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A patient gets 2.1 L of fluid over 19 hours through an IV. The drop factor is 20 gtt/mL.Calculate the drip rate in drops per min
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<span>The rate of infusion is 2.1L/19h or 2100mL/19h (as 1L = 100 mL).
       
To convert 19 hours to minutes we multiply as follows:
   
19 hours = (19 hours) x (60 minutes/1 hour) = 1140 minutes
       
So the rate of infusion becomes:
   
2100mL /1140 min
       
In order to converted mL to drops (gtt) we multiply the rate of infusion with the drop factor to get the drip rate:
       
(2100mL/1140min) x (20 gtt/mL) = 36.8 gtt/min</span>
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