First calculate for the molar mass of the given formula unit, CaCO₃. This can be done by adding up the product when the number of atom is multiplied to its individual molar mass as shown below.
molar mass of CaCO₃ = (1 mol Ca)(40 g Ca/mol Ca) + (1 mol C)(12 g of C/1 mol of C) + (3 mols of O)(16 g O/1 mol O) = 100 g/mol of CaCO₃
Then, divide the given amount of substance by the calculated molar mass.
number of moles = (20 g)(1 mol of CaCO₃/100 g)
number of moles = 0.2 moles of CaCO₃
<em>Answer: 0.2 moles</em>
Answer: 159 grams
Explanation:
Copper (ii) oxide has the chemical formula CuO.
Now given that:
Mass of CuO in grams = ? (let unknown value be Z)
Number of moles = 2.00 moles
Molar mass of CuO = ?
For the molar mass of CuO: Atomic mass of Copper = 63.5g ; Oxygen = 16g
= 63.5g + 16g
= 79.5 g/mol
Apply the formula:
Number of molecules = (mass in grams/molar mass)
2.00 moles = (Z / 79.5 g/mol)
Z = 79.5 g/mol x 2.00 moles
Z = 159g
Thus, there are 159 grams in 2.00 moles of copper (ii) oxide
Answer:
0.292 g/mL.
Explanation:
From the question given above, the following data were obtained:
Mass of object = 28.1 g
Volume of object = 96.2 mL
Density of object =..?
Density of an object is simply defined as the mass of the object per unit volume of the object. Mathematically, it can be expressed as:
Density = mass / volume
With the above formula, we can obtain the density of the object as follow:
Mass of object = 28.1 g
Volume of object = 96.2 mL
Density of object =..?
Density = mass / volume
Density = 28.1 / 96.2
Density of object = 0.292 g/mL
Thus the density of the object is 0.292 g/mL
Answer:
fruit cake is the right ans
Answer:
A. Clear and Sunny
Explanation:
The answer would be A. Clear and Sunny. A high pressure system occurs where the air mass above the Earth is denser than in surrounding areas, and therefore exerts a higher force or pressure. They usually happen with li9ght winds. Using process of elimination, it could not be D. Warm and Stormy, C. Cloudy and Rainy, or B. Cold and Stormy because all have wet climates.