Answer: 3d
Explantion:
1) Period 4 contains the elements with atomic numbers 19 through 36.
2) The elements with atomic numbers 19 (K) and 20 (Ca) fill the orbital 4s.
3) After that, as Aufbau's rule may help you to remember, the energy of the orbitals 3d is lower than the energy of the orbtitals 4p. So, the element 21 (Sc) start fillind the orbital 3d.
There are ten 3d orbitals, so the elements 21 through 30 fill the 3d orbitals.
Those elements, called transition metals are: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, and Zn.
When the 3d orbitals are full, the next elements in the same period 4, fill the six 4p orbitals.
The given compound is 
Each formula unit of has 2 mol Potassium (K) atoms, 2 mol Chromium (Cr) atoms and 7 mol Oxygen (O) atoms.
Atomic weight of potassium is 39.0983 amu
Atomic weight of Chromium is 51.9961 amu
Atomic weight of Oxygen is 15.9994 amu.
Therefore the formula weight of 
= 294.1846 amu
The lines are violet and blue respectively.
<h3>What is the energy?</h3>
We know that the energy of the photon could be obtained by the use of the equation;
E = hf
E = energy
h = Plank's constant
f = frequency
For the first line;
E = 6.6 * 10^-34 Js * 3.45 x 10^14 Hz = 2.3 * 10^-19 J
Given that;
E = hc/λ
λ = hc/E
λ = 6.6 * 10^-34 * 3 * 10^8/2.3 * 10^-19
λ = 8.61 * 10^-7 m or 861 nm
The color is violet
For the second line;
E = 6.6 * 10^-34 Js * 6.53 xx 10^14 Hz
E = 4.3 * 10^-19 J
E = hc/λ
λ = hc/E
λ = 6.6 * 10^-34 * 3 * 10^8/4.3 * 10^-19
λ = 4.60 * 10^- 7 m or 460 nm
The color is blue
Learn more about emission spectrum:brainly.com/question/13537021
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Answer:
63.02 g.
Explanation:
- Na reacts with Cl₂ according to the balanced equation:
<em>2Na + Cl₂ → 2NaCl,</em>
It is clear that 2 mole of Na react with 1 mole of Cl₂ to produce 2 moles of NaCl.
- Firstly, we need to calculate the no. of moles of Na and Cl₂:
no. of moles of Na = (mass/atomic mass) = (19.0 g/22.9897 g/mol) = 0.826 g.
no. of moles of Cl₂ = (mass/atomic mass) = (34.0 g/70.906 g/mol) = 0.48 g.
- From the stichiometry, Na reacts with Cl₂ with a molar ratio (2:1).
<em>So, 0.826 mol of Na "the limiting reactant" reacts completely with 0.413 mol of Cl₂ "left over reactant".</em>
The no. of moles of Cl₂ remained after the reaction = 0.48 mol - 0.413 mol = 0.067 mol.
∴ The mass of Cl₂ remained after the reaction = (no. of moles of Cl₂ remained after the reaction)(molar mass of Cl₂) = (0.067 mol)(70.906 g/mol) = 4.75 g.
- To get the no. of grams of produced NaCl:
<u><em>using cross multiplication:</em></u>
2 mol of Na produce → 2 mol of NaCl, from the stichiometry.
∴ 0.826 mol of Na produce → 0.826 mol of NaCl.
∴ The mass of NaCl produced after the reaction = (no. of moles of NaCl)(molar mass of NaCl) = (0.826 mol)(58.44 g/mol) = 48.27 g.
∴ The total weight of the glass vial containing the final product = the weight of the glass vial + the weight of the remaining Cl₂ + the weight of the produced NaCl = 10.0 g + 4.75 g + 48.27 g = 63.02 g.
Nacl aq. is the best conductor of electricity as in aques state the ions become lose or mobile which are free to move and conduct electricity.
