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 question is incomplete. The complete question is :
C. Balance these fossil-fuel combustion reactions. (1 point)
C8H18(g) + 12.5O2(g) → ____CO2(g) + 9H2O(g) + heat
CH4(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
C3H8(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
C6H6(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
Solution :
C8H18(g) + 12.5O2(g) → __8__CO2(g) + 9H2O(g) + heat
When 1 part of octane reacts with 12.5 parts of oxygen, it gives 8 parts of carbon dioxide and 9 parts of water along with liberation of energy.
CH4(g) + __2__O2(g) → __1__CO2(g) + __2__H2O(g) + heat
When 1 part of methane reacts with 2 parts of oxygen, it gives 1 part of carbon dioxide and 2 parts of water along with liberation of energy.
C3H8(g) + __5__O2(g) → __3__CO2(g) + __4__H2O(g) + heat
When 1 part of propane reacts with 5 parts of oxygen, it gives 3 part of carbon dioxide and 4 parts of water along with liberation of energy.
C6H6(g) + __1/2__O2(g) → __6__CO2(g) + __3__H2O(g) + heat
When 1 part of propane reacts with 1/2 parts of oxygen, it gives 6 part of carbon dioxide and 3 parts of water along with liberation of energy.
Answer:
The best equipment would be the graduated cylinder. Why?
Firstly, the smallest marking on the graduated cylinder is 2 mL, while on all the others the smallest marking is way above that, like 25 mL and 100 mL.
Without even going into the details, we can first rule out the volumetric flask, since its smallest marking is 100 mL and even that is already bigger than our sample size, hence we would have no markings to accurately measure out 29 mL of our sample had we used the volumetric flask.
Next to be ruled out would be the Erlenmeyer flask, as you can see in the image, it only has three marking, and as the smallest marking is 25 mL, each marking is at least 25 mL, and even so far as going up to 50 mL. This cannot let us accurately measure 29 mL out at all, due to the markings being way too big to do that. Hence, the Erlenmeyer flask is ruled out.
Finally, the beaker seems to be a worthy candidate! Unfortunately, for the same reason as the Erlenmeyer flask, as you can see in the image each marking represents 10 mL. We cannot measure 9 mL in the beaker accurately, and hence the beaker is ruled too.
We are left with the graduated cylinder, and that is our answer.
Explanation:
Hope this helped!
<span>The correct answer is option 1. It is the carbon (C) atom that can bond with each other to form ring and chain structures in compounds. These ring and chain structures and compounds are called organic compounds. Chain structures include alkanes, alkenes, alkynes. Ring structures include cyclic hydrocarbons and aromatic hydrocarbons.</span>
A theory must be supported by many different experiments.
