Answer : Malik's experiment included to create artificial atmospheric oxygen; in order to perform experiment on red-legged grasshopper for growing it larger than normal, the part of recreating or mimicking the artificial oxygen for the experiment with additional 10% increase seems to be bit difficult and this may create a problem in defining the accuracy and precision for the experiment.
I believe it is a large mass of warm air rises, replacing the cooler air above
Answer:
Error of parallax it usually occurs when the cylinder is above or below the eye level,and thus resulting in differences In reading the millimeters
Answer:
(a) 1.95 × 10⁴ g
(b) 1.95 × 10⁷ mg
(c) 1.95 × 10¹⁰ μg
Explanation:
A dog has a mass of 19.5 kg.
<em>(a) What is the dog's mass in grams?</em>
1 kilogram is equal to 10³ grams. The mass of the dog in grams is:
19.5 kg × (10³ g/1 kg) = 1.95 × 10⁴ g
<em>(b) What is the dog's mass in milligrams?</em>
1 gram is equal to 10³ milligrams. The mass of the dog in milligrams is:
1.95 × 10⁴ g × (10³ mg/1 g) = 1.95 × 10⁷ mg
<em>(c) What is the dog's mass in micrograms?</em>
1 gram is equal to 10⁶ micrograms. The mass of the dog in micrograms is:
1.95 × 10⁴ g × (10⁶ μg/1 g) = 1.95 × 10¹⁰ μg
Answer:
11.2 M → [HCl]
Explanation:
Solution density = Solution mass / Solution volume
35.38 % by mass, is the same to say 35.38 g of solute in 100 g of solution.
Let's determine the moles of our solute, HCl
35.38 g . 1 mol/36.45 g = 0.970 moles
Let's replace the data in solution density formula
1.161 g/mL = 100 g / Solution volume
Solution volume = 100 g / 1.161 g/mL → 86.1 mL
Let's convert the volume to L → 86.1 mL . 1L / 1000 mL = 0.0861 L
Molarity (M) → mol/L = 0.970 mol / 0.0861 L → 11.2 M