1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.
Let's consider the reaction for the combustion of Mg.
Mg + 1/2 O₂ ⇒ MgO
1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:
We can calculate the mass percent of O in MgO using the following expression.
You can learn more about mass percent here: brainly.com/question/14990953
Answer: n = 3.0 moles
V = 60.0 L
T = 400 K
From PV = nRT, you can find P
P = nRT/V = (3.0 mol)(0.0821 L-atm/K-mol)(400 K)/60.0L
P = 1.642 atm = 1.6 atm (to 2 significant figures)
Answer:
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Explanation:
Answer: A more electronegative atom will have more attraction to the electrons in a chemical bond.
Explanation:
An atom that is able to attract electrons or shared pair of electrons more towards itself is called an electronegative atom.
For example, fluorine is the most electronegative atom.
Due to its high electronegativity it is able to attract an electropositive atom like H towards itself. As a result, both fluorine and hydrogen will acquire stability by sharing of electrons.
Thus, we can conclude that a more electronegative atom will have more attraction to the electrons in a chemical bond.
First step: convert 22.4% into fraction
that is 22.4/100 which is equivalent to 0.224
The mass of Cucl2 contained in 75.85 of 22.4% by mass of solution of CUcl2 in water is therefore,
0.224 x75.85=16.99g
alternatively
75.85 --->100%
? 22.4%
by cross multiplication =(22.4 x 75.85)/100 =16.99g
