The atomic number of an atom is determined by the number of protons it has..
It is also the whole number shown on the periodic table
Answer:
48%
Explanation:
Based on Gay-Lussac's law, the pressure is directly proportional to the temperature. To solve this question we must assume the temperature increases and all CO2 remains without reaction. The equation is:
P1T2 = P2T1
<em>Where Pis pressure and T absolute temperature of 1, initial state and 2, final state of the gas:</em>
P1 = 10.0atm
T2 = 1420K
P2 = ?
T1 = 730K
P2 = 10.0atm*1420K / 730K
P2 = 19.45 atm
The CO2 reacts as follows:
2CO2 → 2CO+ O2
Where 2 moles of gas react producing 3 moles of gas
Assuming the 100% of CO2 react, the pressure will be:
19.45atm * (3mol / 2mol) = 29.175atm
As the pressure rises just to 24.1atm the moles that react are:
24.1atm * (2mol / 19.45atm) = 2.48 moles of gas are present
The increase in moles is of 0.48 moles, a 100% express an increase of 1mol. The mole percent that descomposes is:
0.48mol / 1mol * 100 = 48%
Two changes would make this reaction reactant-favored
C. Increasing the temperature
D. Reducing the pressure
<h3>Further explanation</h3>
Given
Reaction
2H₂ + O₂ ⇒ 2H₂0 + energy
Required
Two changes would make this reaction reactant-favored
Solution
The formation of H₂O is an exothermic reaction (releases heat)
If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)
While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient
in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2
As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)
And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater
Answer:
magnesium metal melts = physical change
magnesium metal ignites = chemical change
Explanation:
<em>Physical changes</em> are those in which the identity of the subtance <u>remains unaltered</u>. No new compounds are formed. They involve generally changes in <u>agreggation states of matter</u>: solid, liquid or gas. The first experiment, in which magnesium metal melts is a physical change because it only changes the state of matter, from solid to liquid, but it is still magnesium metal.
Conversely, <em>chemical changes</em> involve atoms combinations to form new compounds. The second experiment, in which magnesium metal ignites, is a chemical change. After the change, magnesium metal is no longer the metal but a metal oxide.
In an ionic compound the atoms are linked via ionic bonds. These are formed by the transfer of electrons from one atom to the other. The atom that loses electrons gains a positive charge whereas the atom that accepts electrons gains a negative. This happens in accordance with the octet rule wherein each atom is surrounded by 8 electrons
In the given example:
The valence electron configuration of Iodine (I) = 5s²5p⁵
It needs only one electron to complete its octet.
In the given options:
K = 4s¹
C = 2s²2p²
Cl = 3s²3p⁵
P = 3s²3p³
Thus K can donate its valence electron to Iodine. As a result K, will gain a stable noble gas configuration of argon while iodine would gain an octet. This would also balance the charges as K⁺I⁻ creating a neutral molecule.
Ans: Potassium (K)
