Answer:
E. potassium (K) and bromine (Br)
Explanation:
An ionic bond is formed between compounds with a large electronegativity difference between them. It is usually between a metal and non-metal.
- Potassium is a true metal found in group 1 on the periodic table.
- Bromine is a highly electronegative non-metal which is a halogen.
- Potassium will lose one of its electrons which will be gained by the Bromine.
- The electrostatic attraction between the two species will cause the ionic bond to form.
- The ability of one specie willing to lose electron and the other gaining, is the main bed rock of ionic bonding.
Answer:
0.0123 moles
Explanation:
Concentration = Moles / Volume of solution
or you can rearrange the formula to get
Moles = concentration (moles/liter) x volume of solution (liter)
First convert your volume to L instead of mL. 35mL = 0.035L
moles = 0.350 moles/liter x 0.035 liter (liters cancel out)
moles = 0.0123
Answer:
Ecosystem
Explanation:
I got the answer from the Mcgrawhill textbooks.
Answer:
2.893 x 10⁻³ mol NaOH
[HCOOH] = 0.5786 mol/L
Explanation:
The balanced reaction equation is:
HCOOH + NaOH ⇒ NaHCOO + H₂O
At the endpoint in the titration, the amount of base added is just enough to react with all the formic acid present. So first we will calculate the moles of base added and use the molar ratio from the reaction equation to find the moles of formic acid that must have been present. Then we can find the concentration of formic acid.
The moles of base added is calculated as follows:
n = CV = (0.1088 mol/L)(26.59 mL) = 2.892992 mmol NaOH
Extra significant figures are kept to avoid round-off errors.
Now we relate the amount of NaOH to the amount of HCOOH through the molar ratio of 1:1.
(2.892992 mmol NaOH)(1 HCOOH/1 NaOH) = 2.892992 mmol HCOOH
The concentration of HCOOH to the correct number of significant figures is then calculated as follows:
C = n/V = (2.892992 mmol) / (5.00 mL) = 0.5786 mol/L
The question also asks to calculate the moles of base, so we convert millimoles to moles:
(2.892992 mmol NaOH)(1 mol/1000 mmol) = 2.893 x 10⁻³ mol NaOH
