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
Hi.
I did some digging and I think I found what you're looking for.
I found this on Q(uizlet)
basic or acidic conditions and the reactants must be heated.
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Answer:
D, Mixture
Explanation:
Mixtures can be separated physically because that's just how the world works
Answer:
2.64 M
Explanation:
To find the molarity, you need to (1) convert grams to moles (via molar mass), then (2) convert mL to L, and then (3) calculate the molarity (via molarity ratio). The final answer should have 3 sig figs to match the sigs figs of the given values.
(Step 1)
Molar Mass (NH₄NO₃): 2(14.007 g/mol) + 4(1.008 g/mol) + 3(15.998 g/mol)
Molar Mass (NH₄NO₃): 80.04 g/mol
66.5 grams NH₄NO₃ 1 mole
--------------------------------- x ---------------------- = 0.831 moles NH₄NO₃
80.04 grams
(Step 2)
1,000 mL = 1 L
315 mL 1 L
-------------- x ------------------ = 0.315 L
1,000 mL
(Step 3)
Molarity = moles / volume
Molarity = 0.831 moles / 0.315 L
Molarity = 2.64 M
