Answer:
Explanation:
To answer this question successfully, we need to remember that atoms are neutral species, since the number of protons, the positively charged particles, is equal to the number of electrons, the negatively charged particles. That said, we may firstly find an atom which has 3 electrons (and, as a result, 3 protons, as it should be neutral).
The number of protons is equal to the atomic number of an element. We firstly may have an atom with 3 protons and 3 electrons (atomic number of 3, this is Li).
Similarly, we may take the atomic number of 4, beryllium, and remove 1 electron from it. Upon removing an electron, it would become beryllium cation, 
.
We may use the same logic going forward and taking the atomic number of 5. This is boron. In this case, we need to remove 2 electrons to have a total of 3 electrons. Removal of 2 electrons would yield a +2-charged cation: 
.
Answer:
∆H > 0
∆Srxn <0
∆G >0
∆Suniverse <0
Explanation:
We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.
Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.
The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.
Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.
Answer:
Explanation:
Sr(OH)₂.+ 2HCl = SrCl₂ + 2H₂O
Moles of HCl in 28mL of .10 M HCl = .028 x .1 = .0028 moles .
Moles of Sr(OH)₂ in 60mL of .10 M Sr(OH)₂ = .060 x .1 = .0060 moles
2 moles of HCl reacts with 1 mole of Sr(OH)₂
.0028 moles of HCl reacts with .0014 mole of Sr(OH)₂
moles of Sr(OH)₂ remaining = .0060 - .0014 = .0046 moles .
Sr(OH)₂ = Sr⁺ + 2OH⁻
1 mole 2 mole
.0046 .0092
Total volume of solution = 88 mL .
88 mL of solution contains .0092 moles of OH⁻
concentration of OH⁻ = .0092 / .088
= .1045 M .
Answer:
a negative charge
Please Mark Brainliest If This Helped!
Answer:
a.3-5 is the modal class.
