Answer:
2.51 x 1023 atoms C
Explanation:
MW C = 12.011
5g x (1mol/12.011g) =0.416 mol
0.416 mol x (6.022x10^23) = 2.51 x 1023 atoms C
Answer:
pH = 12.8
Explanation:
HF + NaOH → F⁻ + Na⁺ + H₂O
<em>1 mole of HF reacts with 1 mole of NaOH</em>
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Initial moles of HF and NaOH are:
HF = 0.018L × (0.308mol / L) = 5.544x10⁻³mol HF
NaOH = 0.023L × (0.361mol / L) = 8.303x10⁻³mol NaOH
That means moles of NaOH remains after reaction are:
8.303x10⁻³mol - 5.544x10⁻³mol = <em>2.759x10⁻³moles NaOH</em>
Total volume is 18.0mL + 23.0mL = 41.0mL = 0.0410L
Molar concentration of NaOH is
2.759x10⁻³moles NaOH / 0.0410L = 0.0673M = [OH⁻]
pOH = - log [OH⁻] = 1.17
As pH = 14 - pOH
<em>pH = 12.8</em>
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Answer:
True; When one side of a molecule is electronegative (δ-) and the other side of the
molecule is electropositive (δ+), it is said to have a dipole moment.
Explanation:
A dipole moment exists in a molecule as a result of differences in the electronegativity values between the atoms of the elements involved in the chemical bonding.
When a strogly electronegative atom such as oxygen or chlorine is chemically bonded to a less electronegative or an electropositive atom such as hydrogen, there is an uneven sharing of the electrons involved in the bonding. The more electronegative atoms tends to draw the shared electrons mostly to themselves. This induces a partially negative charge (δ-) on them while leaving the electropositive atoms with a partially positive charge (δ+).
Water is an example of a molecule having a dipole moment. The oxygen atoms are more electronegative than hydrogen and as such draw the shared electrons to themselves more, inducing a partial positive charge (δ+) on the hydrogen atoms while they themselves develop a partial negative charge (δ-).
<h2><u>Answer:</u></h2>
The correct answer is A) 1.04 mol Cu
{65.8 g / 63.55 g/mol}
= 1.04 mol Cu
Explanation:
In 63.55 g of copper metal there are 1 m
o
l of C
u atoms. By dividing the mass of Cu and molar mass, we can easily get the number of moles.
The correct answer is C.
Most of the time, double replacements produce one product that is soluble and one that is insoluble
This results in a precipitate within a liquid or aqueous solution
Hope this helps
