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.

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The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L
From the question given above, the following data were obtained:
Initial pressure (P₁) = 8.5 atm
Initial volume (V₁) = 24 L
Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
Final pressure (P₂) = 13.5 atm
Final temperature (T₂) = 15 °C = 15 + 273 = 288 K
<h3>Final volume (V₂) =? </h3>
- The final volume of the gas can be obtained by using the combined gas equation as illustrated below:

Cross multiply
298 × 13.5 × V₂ = 204 × 288
4023 × V₂ = 58752
Divide both side by 4023

<h3>V₂ = 15 L </h3>
Therefore, the final volume of the gas is 15 L
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Answer:
It bonds with the added H+ or OH in solution.
Explanation:
Answer:
Option A. 70.0 KPa.
Explanation:
Data obtained from the question include:
Pressure (torr) = 525.4 torr
Pressure (kPa) =?
The pressure expressed in torr can be converted kPa as shown below:
760 torr = 101.325 KPa
Therefore,
525.4 torr = (525.4 x 101.325) / 760 = 70.0 KPa.
Therefore, 525.4 torr is equivalent to 70.0 KPa.