Given :
Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .
To Find :
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .
Solution :
By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
So , volume of solution does not matter .
Moles of oxygen ,
.
Now , molecule of CO contains 1 mole of C .
So , moles of C is also 0.167 mole .
Mass of carbon ,
.
Therefore , mass of carbon is 2 grams .
Hence , this is the required solution .
The normal range of creatinine in human blood is between 0.50 mg/dL and 1.1 mg/dL. The patient's blood has a concentration of 0.0082 g/L. Let's convert that value into mg/dL.
We kwnot that there are 1000 mg in 1 g. And there are 10 dL in 1 L. We have to use those conversions.
1000 mg = 1 g 10 dL = 1 L
0.0082 g/L = 0.0082 g/L * 1000 mg/g = 8.2 mg/L * 1 L/ (10 dL) = 0.82 mg/dL
0.0082 g/L = 0.82 mg/dL
0.50 mg/dL < 0.82 mg/dL < 1.1 mg/dL
Answer: The concentration of creatinine = 0.82 mg/dL. It is in the normal range.
The electron bit would be 2 and you could colour in 4 boxes:) I had the same question as this and I got it correct