Answer:
placing the reactants on a hot plate
Explanation:
If the temperature goes up, the reaction rate will increase. Because the particle will move faster and makes the kinetic energy larger.
Answer:
a. K⁺
b. Se²⁻
Explanation:
Isoelectronic species are those that have the same number of electrons.
a.
Ar has 18 electrons. K has 19 electrons, so when it loses 1 electron to form K⁺, it is isoelectronic with Ar.
b.
Kr has 36 electrons. Se has 34 electrons, so when it gains 2 electrons to form Se²⁻, it is isoelectronic with Kr.
Answer:
-266 kJ
Explanation:
There is some info missing. I think this is the original question.
<em>A mixture of nitrogen and neon gas is expanded from a volume of 53.0 L to a volume of 90.0 L, while the pressure is held constant at 71.0 atm. Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.</em>
<em />
Given data
Initial volume = 53.0 L
Final volume = 90.0 L
Pressure = 71.0 atm
We can find the work (w) associated with the expansion of the gaseous mixture using the following expression.
w = - P × ΔV = - 71.0 atm × (90.0 L - 53.0 L) = -2.63 × 10³ atm . L
w = -2.63 × 10³ atm . L × (101.3 J/ 1 atm . L) = -2.66 × 10⁵ J = -266 kJ
The negative sign means that the system does work on the surroundings.
Answer:
[Ar] 3d10 4s1
Explanation:
The correct electronic configuration of copper is [Ar] 3d10 4s1
Copper has atomic number 29 and due to the stability of half filled or fully filled orbitals or shells, the electrons from the 4s jumps to the 3d and makes the 3d shell contain 10 electrons.
This is what I mean:
Cu = Ar 4s2 3d10 is the expected configuration of copper when we follow the principle of filling the various orbitals that is the s, p, d, f orbitals.
But because we write 3d before writing 4s, we have Ar 3d10 4s2. Instead of this configuration becoming the correct one, an electron from the 4s orbital jumps to the 3d orbital to complete the orbital giving the electrons in the 3d orbital 10.
So therefore the correct configuration is Ar 3d10 4s1
<span>the solvent, hope this helps</span>
