Actually Rb or Rubidium in zero state has the following
electron configuration:
<span>1s22s2</span><span>2p6</span><span>3s2</span><span>3p63d10</span><span>4s2</span><span>4p65s1</span>
However we can see that the ion has a 1 positive charge,
which means that it lacks 1 electron, therefore the answer from the choices is:
<span>d.
rb+: 1s22s22p63s23p64s23d104p6</span>
Answer:
The cost would be $1.68
Explanation:
I found the answer by multiplying
.67x or .67lb with 2.5, getting $1.68
Answer:
<u><em>Explain both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigon planar; the carbon atom exhibits sp2 hybridization. Two of the sp2 orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining sp2 hybrid orbital forms a σ bond to the oxygen atom. The hybridized p orbital on the carbon atom in the carbonyl group overlaps a p orbital on the oxygen atom to form the π bond in the double bond.ation .</em></u>
Data Given:
Pressure = P = ?
Volume = V = 3.0 L
Temperature = T = 115 °C + 273 = 388 K
Mass = m = 75.0 g
M.mass = M = 44 g/mol
Solution:
Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ------ (1)
Calculating Moles,
n = m / M
n = 75.0 g / 44 g.mol⁻¹
n = 1.704 mol
Putting Values in Eq. 1,
P = (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L
P = 18.08 atm
(3) The concen<span>tration of the solution remains constant.</span>
