PV = nRT
Pressure times volume = number of moles times ideal gas constant times temperature
50mL = .05L
1 atm (.05) = n (.0821) (273)
.05 = n (22.4133)
.022 = n (number of moles)
6.022E23 molecules = 1 mole
.022 x 6.022E23 = 1.325E23 molecules
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: I believe it's C
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Can you please make my answer brainly
answer:an arrangement of elements in columns, based on a set of properties that repeat from row to row
Explanation:
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Club soda so A, you just gotta match the color that it says for it to the chart and it’ll show the identity the ph which is four I believe It said