Answer is: an oxybromate compound is KBrO₄ (x = 4).
ω(Br) = 43.66% ÷ 100%.
ω(Br) = 0.4366; mass percentage of bromine.
If we take 100 grams of compound:
m(Br) = ω(Br) · 100 g.
m(Br) = 0.4366 · 100 g.
m(Br) = 43.66 g; mass of bromine.
n(Br) = m(Br) ÷ M(Br).
n(Br) = 43.66 g ÷ 79.9 g/mol,
n(Br) = 0.55 mol; amoun of bromine.
From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).
m(K) = 0.55 mol · 39.1 g/mol.
m(K) = 21.365 g; mass of potassium in the compound.
m(O) = 100 g - 21.365 g - 43.66 g.
m(O) =34.97 g; mass of oxygen.
n(O) = 34.97 g ÷ 16 g/mol.
n(O) = 2.185 mol.
n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.
n(K) : n(Br) : n(O) = 1 : 1 : 4.
As we know that
P.E. = mgh
where,
P.E. = Potential energy of the object =?
m= mass of object= 3kg
g= acceleration due to gravity = 9.8 ms^-2
h = height between object and animal = 0 m
Then
P.E. = 3× 9.8 × 0 = 0 Joules or 0J
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The missing components in the table to the right are indicated with orange letters. Use the periodic table in the tools bar and this link Web Elements to fill in the corresponding values. A B C D E F G. 2. See answers. Log in to add ... F = 737.7kJ/mol. G = 495.8kJ/mol. Explanation: We are asked some of the ...
2 answers
Answer:
To have the electronic configuration equal to 1s²2s²2p⁶3s²3p⁶4s²3d⁷, the chemical element must have an electrical charge equal to 27, that is, it must have 27 electrons, such as Cobalt (Co), for example.
Explanation:
The electronic configuration shown in the question above is known as the Linus Pauling distribution and represents the energy sub-levels that an electrically charged atom can have in relation to the amount of electrons it has.
The layers sub-levels are presented in the following order 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹º 4p⁶ 5s² 4d¹º 5p⁶ 6s² 4f14 5d¹º 6p⁶ 7s² 5f14 6d¹º 7p⁶. Where the small numbers represent the number of electrons in each sub-level and the large numbers represent the layers of electronic distribution.
Accordingly, we can see that an atom that has the configuration 1s²2s²2p⁶3s²3p⁶4s²3d⁷ has 27 electrons, like Cobalt.
