Apatient is receiving 3000 ml/day of a solution that con-
tains 5 g of dextrose (glucose) per 100 ml of solution. if
glucose provides 4 kcal/g of energy, how many kilocalo-
ries per day is the patient receiving from the glucose?

600 kCal/ day

Explanation:

You can solve this type of problems using proportions, with the rule of three.

Follow these steps:

1. Analyze the giving data

2. Organize it

3.Write the proportions  

4. Clear the value you are looking for

5. Solve the equations by multiplying or dividing

Now for the exercise:

You have the following data:

3000 mL Solution per 1 day > 3000 mL S

The concentration of the solution is  

5 g of dextrose per 100 mL  Solution > 5g D - 100 mL S

The glucose provides some specific energy

4 kCal per 1 g of dextrose > 4kCal - 1 g D

First find the total grams of  dextrose (glucose) given in one day.

3000 mL S - x g D?

 100 mL S - 5 g D

The grams of glucose given in one day

xg D? = 3000 mL S(5 g D)/100 mL S  

you also can cancel the units mL S

xg D? = 150 g D.

The grams given in one day of glucose are 150 g /day

Now find the total Energy in those grams:

4 kCal -      1 g D

x kCal?-150 g D

Clear x kCal?

x kCal? = 4kCal (150g D)/(1g D)

x kCal? = 600 kCal  remember is for one day!

so the answer is

The patient is receiving 600 kCal/day

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

→

a:

by the combination of an unlimited number of elements