1.65g MgO = 1g Mg
1.65 - 1 = 0.65 g of O in MgO
solve it using proportion:
1g Mg / 0.65g O = x (g) Mg / 16g O
or 1 / 0.65 = x / 16
24.6 g is the answer.
if 1 gram of oxygen requires 1.65 grams of Mg
then 16 grams of oxygen will require 16 ( 1.65) or 26.4 grams.
Answer: potential.
Chemical energy is the energy provided by a chemical reaction.
Kinetic energy is the energy due to the speed.
Potential energy is the energy due to the position. For example, an object on the top of a mountain, has the possibility to perform work if it falls.
Electromagnetic energy. is propagated by waves: radio waves, infrared radiation, microwaves, etc.
Answer
pH=8.5414
Procedure
The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.
pH = pKa + log₁₀ ([A⁻] / [HA])
Where
pH = acidity of a buffer solution
pKa = negative logarithm of Ka
Ka =acid disassociation constant
[HA]= concentration of an acid
[A⁻]= concentration of conjugate base
First, calculate the pKa
pKa=-log₁₀(Ka)= 8.6383
Then use the equation to get the pH (in this case the acid is HBrO)
Answer:
The pressure of the gas would be 3.06 atm
Explanation:
Amonton's law states that the pressure is directly proportional to the absolute temperature of a gas under constant volume. The equation is:
P1 / T1 = P2 / T2
<em>Where P1 is the initial pressure = 3.16atm</em>
<em>T1 is initial absolute temperature = 273.15 + 32.2°C = 305.35K</em>
<em>P2 is our incognite</em>
<em>And T2 is = 273.15 + 22.9°C = 296.05K</em>
<em />
Replacing:
3.16atm / 305.35K = P2 / 296.05K
3.06 atm = P2
<h3>The pressure of the gas would be 3.06 atm</h3>
Solids, liquids, gases, and plasma are all matter. When all atoms that make up a substance are the same, then that substance is an element. Elements made of only one kind of atom. Because of this, elements are called "pure" substances.
hope this helps :)
