Answer:
1.332 Molar is the molarity of the glucose solution.
Explanation:
Molarity of the solution is the moles of compound in 1 Liter solutions.
Mass of glucose = 60.00 g
Molar mass of glucose = 180.2 g/mol
Volume of the solution = V = 250.0 mL = 0.250 L
(1 mL = 0.001 L)
1.332 Molar is the molarity of the glucose solution.
Calcium Phospate formula: Ca3(PO4)2
Atomic weight:
Ca = 40 ; P = 31 ; O = 16
Ca3 = 40 * 3 = 120
P = 31
O4 = 16 * 4 = 64
(PO4)2 = (31 + 64) * 2 = 95 * 2 = 190
Ca3(PO4)2 = 120 + 190 = 310 g/mol
31/310 = 10% P in the calcium phospate
1.00kg * 1000 g/kg = 1000 g of phosphorus.
1000 g / 10% = 10,000 g of calcium phosphate
10,000 g / 58.5% = 17,094 grams of ore.
The minimum mass of the ore should be 17,094 grams or 17.094 kg.
It is a scientific hypothesis. A scientific hypothesis must be testable, however there is a significantly more grounded necessity that a testable speculation must meet before it can truly be viewed as logical. This foundation comes essentially from crafted by the rationalist of science Karl Popper, and is called "falsifiability".
1 teaspoon contains <span>76.002146 drops
To know how many drops do 0.171 teaspoon contains, we can simply use cross multiplication as follows:
number of drops = (0.171 x </span><span>76.002146) / 1 = 12.9963 drops</span>
Answer:
CaCl2 + 2NaOH -> 2NaCl + Ca(OH)2
Explanation:
