Answer:
2Rb(s) + Sr^+(aq) → 2Rb^+ (aq) + Sr(s)
Explanation:
Rubidium has a more negative reduction potential (-2.98 V) compared to strontium (-2.89 V).
Hence, in a redox reaction involving rubidium and strontium, rubidium will be oxidized while strontium is reduced.
The balanced redox reaction equation is obtained from;
Oxidation half equation;
2Rb(s) ---->2Rb^+(aq) + 2e
Reduction half equation;
Sr^2+(aq) + 2e ----> Sr(s)
Overall reaction equation;
2Rb(s) + Sr^+(aq) → 2Rb^+ (aq) + Sr(s)
Answer:
4.27 L
Explanation:
STP = 0 C and 1 atm
P1V1/T1 = P2V2/T2 where T is in Kelvin
P1 V1 / T1 *T2/P2 = V2
.469 * 5 / ( 273.15-123) * 273.15/1 = V2 = 4.27 L
Answer:
17.76g
Explanation:
We need to write a balanced chemical equation for the reaction:
2Al(OH)3 + 3Ca(NO3)2 ——> 3Ca(OH)2 + 2Al(NO3)3
In the reaction above, it can be seen that 2 moles of aluminum hydroxide yielded 3 moles of calcium hydroxide. This is the theoretical viewpoint.
Now we need to know what actually happened. We need to calculate the actual number of moles of aluminum hydroxide reacted l. We can get this by dividing the mass by the molar mass.
The molar mass of aluminum hydroxide is 27+ 3( 16+1)
= 27 + 51 = 78g/mol
The number of moles is thus: 12.55/78 = 0.16 moles
Now if 2 moles of aluminum hydroxide gave 3 moles of calcium hydroxide, 0.16moles will give : (0.16*3)/2 = 0.24moles
Now we can calculate the mass of calcium hydroxide formed. The mass of calcium hydroxide formed is the number of moles multiplied by the molar mass.
The molar mass of calcium hydroxide is; 40 + 2(17) = 74g/mol
The mass is thus =74 * 0.24 = 17.76g
Answer:
0.056 M
Explanation:
To calculate the concentration of h we must use the following formula:
10^-pH
So lets plug in our pH:
10⁻¹.²⁵
When we plug this into our calculator our answer is
0.056 M
Answer:
σ -> 2sp²
π -> 2p
Explanation:
The carbon has valence shell 2s 2p, and, both of them make 3 σ bonds and 1 π bond. The π bond only occurs in multiple bonds.
The σ bonds happen at the hybrids orbitals, which are orbitals formed by the association of the pure orbitals (s, p, d, f). The hybridization occurs to make possible to the atom to do the bonds because the electrons need to be isolated in it.
On the other hand, the π bonds only occur at pure orbitals. The subshell s only has 1 orbital, and the subshell p has 3 orbitals. So, because there are 3 σ bonds, it's necessary 3 hybrids orbitals (1 of s + 2 of p).
The σ bonds happen at the orbital 2sp² and the π bond at the 2p pure orbital.
