Only coniferous trees have thin, waxy, needle-like leaves that drop throughout the year and help to retain water.
<h3>What are coniferous trees?</h3>
Coniferous trees are evergreen trees that produce cones.
In addition, they produce waxy and needle-like leaves that help them conserve water by limiting the evapotranspiration rate as a result of reduced surface area.
Examples of conifers include pine trees and spruces.
More on conifers can be found here: brainly.com/question/23222416
Answer:
1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵
Explanation:
1) Ca:
Its atomic number is 20. So it has 20 protons and 20 electrons.
Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.
So, you have two more electrons (20 - 18 = 2) to distribute.
Those two electrons go the the orbital 4s.
Finally, the electron configuration is [Ar] 4s².
2) Pm
The atomic number of Pm is 61, so it has 61 protons and 61 electrons.
Pm is in the row (period) 6. So, the noble gas before Pm is Xe.
The atomic number of Xe is 54.
Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.
The resultant distribution for Pm is: [Xe]6s² 4f⁵.
Answer: Solution A : ![[H_3O^+]=0.300\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D0.300%5Ctimes%2010%5E%7B-7%7DM)
Solution B : ![[OH^-]=0.107\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.107%5Ctimes%2010%5E%7B-5%7DM)
Solution C : ![[OH^-]=0.177\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.177%5Ctimes%2010%5E%7B-10%7DM)
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.
![pH=-\log[H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH_3O%5E%2B%5D)
![[H_3O^+][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
a. Solution A: ![[OH^-]=3.33\times 10^{-7}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3.33%5Ctimes%2010%5E%7B-7%7DM)
![[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B3.33%5Ctimes%2010%5E%7B-7%7D%7D%3D0.300%5Ctimes%2010%5E%7B-7%7DM)
b. Solution B : ![[H_3O^+]=9.33\times 10^{-9}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D9.33%5Ctimes%2010%5E%7B-9%7DM)
![[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B9.33%5Ctimes%2010%5E%7B-9%7D%7D%3D0.107%5Ctimes%2010%5E%7B-5%7DM)
c. Solution C : ![[H_3O^+]=5.65\times 10^{-4}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D5.65%5Ctimes%2010%5E%7B-4%7DM)
![[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B5.65%5Ctimes%2010%5E%7B-4%7D%7D%3D0.177%5Ctimes%2010%5E%7B-10%7DM)
When HCl reacts with a metal, hydrogen gas will be evolved. To test this gas, insert a burning splinter into the outlet of gas, the flame will be extinguished with a pop sound. This will confirm the gas is hydrogen.
