Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.
We make a proportion out of the word problem
(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose
A) a column
example: earth alkaline metals
Answer:
The first one is 425g of copper
Explanation:
Answer:
K₂CO₃
Explanation:
Given parameters:
Number of moles of K = 0.104mol
Number of moles of C = 0.052mol
Number of moles of O = 0.156mol
Method
From the given parameters, to calculate the empirical formula of the elements K, C and O, we reduce the given moles to the simplest fraction.
Empirical formula is the simplest formula of a compound and it differs from the molecular formula which is the actual formula of a compound.
- Divide the given moles through by the smallest which is C, 0.052mol.
- Then approximate values obtained to the nearest whole number of multiply by a factor to give a whole number ratio.
- This is the empirical formula
Solution
Elements K C O
Number of moles 0.104 0.052 0.156
Dividing by the
smallest 0.104/0.052 0.052/0.052 0.156/0.052
2 1 3
The empirical formula is K₂CO₃
