<h3><u>Answer;</u></h3>
2.4 atm
<h3><u>Explanation;</u></h3>
At constant temperature;
PV = constant
P1 = 1.2 atm
V1 = 0.25 L
P2 = ?
V2 = 0.125 L
But;
P1*V1 = P2*V2
Thus;
P2 = P1*V1/V2
= (1.2 atm)(0.25 L)/(0.125 L)
<u> = 2.4 atm</u>
Answer:
Explanation:
All chemical changes requires a substantial amount of energy. Depending on the type of reaction, the energy of reactants and products differs.
In an exothermic chemical change where heat is always given off to the surroundings, the energy of the product is less than that of reactants.
For endothermic changes, the surrounding becomes colder at the end of the reaction. This implies that the heat energy of the product is higher than that of the reactants.
Answer:
We need 420 cal of heat
Explanation:
Step 1: Data given
Mass of the aluminium = 200.0 grams
Temperature rises with 10.0 °C
Specific heat of aluminium = 0.21 cal/g°C
Step 2: Calculate the amount of heat required
Q =m * c* ΔT
⇒with Q = the amount of heat required= TO BE DETERMINED
⇒with m = the mass of aluminium = 200.0 grams
⇒with c = the specific heat of aluminium = 0.21 cal/g°C
⇒with ΔT = the change of temperature = 10.0°C
Q = 200.0 grams * 0.21 cal/g°C * 10.0 °C
Q = 420 cal
We need 420 cal of heat (option 2 is correct)
Answer is: a. Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Alkaline metals (group 1), in this example rubidium, have lowest ionizations energy and easy remove valence electrons (one electron), they are most reactive metals.
Earth alkaline metals (group 2), in this example strontium, have higher ionization energy than alkaline metals, because they have two valence electrons, they are less reactive.
Rubidium electron configuration: ₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹; one valence electron is 5s¹ orbital.
Strontium electron configuration: ₃₈Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²; two valence electrons is 5s² orbital.
I can't remember any of the weights of the individual elements but here is how you solve it:
Molecular weight of copper + nitrogen + 3 oxygens = molecular weight of the compound.
M = moles / liter
.350 moles / 1 liter
Do .350 moles / liter x the molecular weight (g / mole) of the compound = the answer in g / L
