Darwin used finches in his theory of evolution.
Answer:
Explanation:
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Answer:
1. The electronic configuration of X is: 1s2 2s2 sp6 3s2
2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
3. The formula of the compound form by X and Y is given as: XY
Explanation:
For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:
1s2 2s2 sp6 3s2
To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:
Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below
1s2 2s2 2p4
The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
The formula of the compound form by X and Y is given below :
X^2+ + Y^2- —> XY
Their valency will cancel out thus forming XY
132 g of C , 22 g of H , 176 g of O
132 + 22 + 176 => 330 g <span>of the substance
</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u H = 1.0 u O = 16.0 u
C = 132 / 12.0 => 11 moles
H = 22 / 1.0 => 22 moles
O = 176 / 16.0 => 11 moles
Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1
H = 22 / 11 => 2
O = 11 / 11 => 1
formula empirically <span>is : CH</span>₂O
hope this helps!
Answer:
0.166M
Explanation:
In a neutralization, the acid, H₂SO₄, reacts with a base, KOH, to produce a salt, K₂SO₄ and water. The reaction is:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
To solve this problem, we need to determine moles of H2SO4 and moles of KOH that reacts to find the moles of sulfuric acid that remains after the reaction:
<em>Moles H2SO4:</em>
0.650L * (0.430mol /L) = 0.2795moles H2SO4
<em>Moles KOH:</em>
0.600L * (0.240mol / L) = 0.144 moles KOH
Moles of sulfuric acid that reacts with 0.144 moles of KOH are:
0.144 moles KOH * (1mol H2SO4 / 2 mol KOH) = 0.072 moles of H2SO4 react.
And remain:
0.2795moles H2SO4 - 0.072moles H2SO4 = 0.2075 moles of H2SO4 reamains.
In 0.650L + 0.600L = 1.25L:
Molar concentration of sulfuric acid:
0.2075 moles of H2SO4 / 1.25L =
<h3>0.166M</h3>
