Answer:
So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.
Explanation:
What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.
What are the names of the two magnetic poles?
Ur answer is
North and South
Answer:
10.87 g of Ethyl Butyrate
Solution:
The Balance Chemical Equation is as follow,
H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH → H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O
According to equation,
88.11 g (1 mol) Butanoic Acid produces = 116.16 g (1 mol) Ethyl Butyrate
So,
8.25 g Butanoic Acid will produce = X g of Ethyl Butyrate
Solving for X,
X = (8.25 g × 116.16 g) ÷ 88.11 g
X = 10.87 g of Ethyl Butyrate
Answer:
<h2>Dead & Decaying Organisms</h2>
Explanation:
There are many inorganic components of soil, such as clay. However, soil is mainly comprised of dead and decaying organisms. For example, old, and decaying plants.
<em>PLEASE MARK BRAINLIEST</em>
Elements that exceed the octet rule must have an unoccupied d orbital. I believe.
