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:
109.7178g of H2O
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
2C3H8O + 9O2 —> 6CO2 + 8H2O
Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:
Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.
Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g
Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol
Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g
From the equation,
120.19184g of C3H8O produced 144.12224g of H20.
Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O
The statement of the combined gas law for a fixed amount of gas is,
PV/T = constant
Here, the units of pressure and volume must be consistent and the temperature must be the absolute temperature (Kelvin or Rankine).
0.65 atm is equivalent to 494 mmHg
Using the equation:
(755 x 500) / (27 + 273) = (494 x V) / (-33 + 273)
V = 3396 ml = 3.4 liters
Answer:
The kind of ionic compound formed is MX2.
Explanation:
Element X electron configuration is represented as [core] ns2np5. The group in the periodic table this element belong to is group 7A. The element group is called the halogen family. Element X cannot be stated specifically, because the number is represented with n. Element X will behave as an anions when it react with a metal(cations). Element X has a charge of -1. The element X will gain electron when it bond with a metal. Element X is a non metal . Elements in this group are fluorine, chlorine, bromine, iodine , astatine, and tennessine . The element X have 7 valency electrons.
Element M electronic configuration is represented as [core]ns2. The group in the periodic table this element belong to is group 2A . The element group is called the alkaline earth metals family . Element M will behave as a cation when it bond with a non metal. Element M is a metal , therefore it will likely lose electron to form cations during bonding . The charge of element M is 2+. Element M is positively charged. Elements that belong to this group includes beryllium, magnesium, calcium, strontium, barium and radium. Element M has 2 valency electrons.
The reaction between this 2 ions will likely form an ionic compound . The element M is the cations while the element X is the anions. The element M will lose 2 electron while 2 atoms of element X will gain 2 electrons.Element M will lose 2 electron to attain a stable configuration while 2 atoms of element X will gain a single electron each to attain a stable configuration.
M²+ and F- . This will form MX2 when you cross multiply the charge. The kind of ionic compound formed is MX2.
I think it’s C atomic radius and numbers of unshielded protons
