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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a mixture containing 21.4 g of ice. (0.00 c) and 75.3 g of water (55.3 c) is placed in an insulatated container

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3 years ago

One of relatively few reactions that takes place directly between two solids at room temperature is In this equation, the in Ba(

yuradex [85]

Answer:

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate.

Explanation:

$Ba(OH)_2.8H_2O(s)+NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+8H_2O(l)+NH_3(g)$

The balance chemical equation is :

$Ba(OH)_2.8H_2O(s)+2NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+10H_2O(l)+2NH_3(g)$

Mass of barium hydroxide octahydrate = 6.5 g

Moles of barium hydroxide octahydrate = $\frac{6.5 g}{315 g/mol}=0.020635 mol$

According to reaction, 2 moles of ammonium thiocyanate reacts with1 mole of barium hydroxide octahydrate. The 0.020635 moles of barium hydroxide octahydrate will react with:

$\frac{2}{1}\times 0.020635 mol=0.04127 mol$