The triple beam balance<span> is used to </span>measure<span> masses very precisely; the reading error is 0.05 gram.</span>
Answer:
From the numerical steps highlighted under explanation, the average atomic mass of bromine is 79.91 u
Explanation:
The steps to be taken will involve;
1) Find the number of isotopes of bromine.
2) Identify the atomic mass and relative abundance of each of the isotopes.
3) Multiply the atomic mass of each of the isotopes by their corresponding values relative abundance value.
4) Add the value in step 3 above to get the average atomic mass of bromine.
Now;
Bromine has 2 isotopes namely;
Isotope 1: Atomic mass = 78.92amu and a relative abundance of 50.69%.
Isotope 2: Atomic mass = 80.92amu and a relative abundance of 49.31%.
Using step 3 above, we have;
(78.92 × 50.69%)
And (80.92 × 49.31%)
Using step 4 above, we have;
(78.92 × 50.69%) + (80.92 × 49.31%) ≈ 79.91 u
Answer:
![\boxed{\sf D. \ [Ar] \ 3d5 \ 4s2}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%20D.%20%5C%20%20%5BAr%5D%20%5C%203d5%20%5C%204s2%7D)
Explanation:
Manganese is a group 7 transition metal with an electronic configuration of [Ar] 3d5 4s2.
Answer:
because of heat from the sidewalk
Answer:
4
10
Explanation:
The reaction equation is given as;
Ca(OH)₂ → Ca²⁺ + 2OH⁻
Concentration of Ca(OH)₂ = 5 x 10⁻⁵M
Unknown:
pOH of the solution = ?
pH of the solution = ?
Solution:
Solve for the pOH of this solution using the expression below obtained from the ionic product of water;
pOH = ⁻log₁₀[OH⁻]
Ca(OH)₂ → Ca²⁺ + 2OH⁻
1moldm⁻³ 1moldm⁻³ 2 x 1moldm⁻³
5 x 10⁻⁵moldm⁻³ 5 x 10⁻⁵moldm⁻³ 2( 5 x 10⁻⁵moldm⁻³ )
1 x 10⁻⁴moldm⁻³
Therefore;
pOH = -log₁₀ 1 x 10⁻⁴ = 4
Since
pOH + pH = 14
pH = 14 - 4 = 10