Which contains the greatest number of bromide ions?
a) 100 mL of 1.0 M BaBr2
b) 50 mL of 2.0 M NaBr
c) 20 mL of 2.5 M FeBr3
d) All contain the same number of chloride ions.
How would I go about this question?
If you want a "complete" answer,
1. determine mols of Br for each from
mols = M x L
2. Multiply mols x 6.02E23
3. Compare to determine the one with the greatest.
OR
recognize that all will be multiplied by 6.02E23 which means #1 and #3 will answer the question.
a. is 1.0 M x 0.100 L x 2 Br/1 mol BaBr2
b. is 2.0 M x 0.050 L x 1 Br/1 mol NaBr.
c. is 2.50 M x 0.020 L x 3Br/1 mol FeBr3.
d. is an imaginary answer; there are no Cl ions in the problem.
Technically there are two answers; i.e., one answer is either a,b, or c but d is correct, too, because there are no Cl ions in any of the solutions.
Well, bromide ions are like the party animals of the chemistry world, always hanging out in solutions and causing a ruckus.
To figure out which solution contains the greatest number of bromide ions, we can use some simple math. We need to calculate the moles of bromide ions in each solution.
For option a) 100 mL of 1.0 M BaBr2, we can use the equation: moles = concentration x volume. So, moles = 1.0 M x 0.100 L = 0.100 moles of bromide ions.
For option b) 50 mL of 2.0 M NaBr, moles = 2.0 M x 0.050 L = 0.100 moles of bromide ions.
For option c) 20 mL of 2.5 M FeBr3, moles = 2.5 M x 0.020 L = 0.050 moles of bromide ions.
So, it looks like options a) and b) both have the same number of bromide ions, with 0.100 moles. Option c) has a lower number of bromide ions with 0.050 moles.
Therefore, the greatest number of bromide ions is found in options a) and b). It's bromance at its finest!
To determine which option contains the greatest number of bromide ions, we need to calculate the number of bromide ions in each solution.
The number of bromide ions can be calculated using the formula:
Number of bromide ions = (concentration in moles/L) x (volume in liters)
Let's calculate the number of bromide ions for each option:
a) 100 mL of 1.0 M BaBr2:
Number of bromide ions = (1.0 mol/L) x (0.1 L) = 0.1 moles
b) 50 mL of 2.0 M NaBr:
Number of bromide ions = (2.0 mol/L) x (0.05 L) = 0.1 moles
c) 20 mL of 2.5 M FeBr3:
Number of bromide ions = (2.5 mol/L) x (0.02 L) = 0.05 moles
d) All contain the same number of chloride ions.
Since this option states that all solutions contain the same number of bromide ions, it implies that each option contains an equal amount of bromide ions as calculated above, which is not the case.
Therefore, the option with the greatest number of bromide ions is a) 100 mL of 1.0 M BaBr2, with 0.1 moles of bromide ions.
To determine which solution contains the greatest number of bromide ions, we need to compare the number of moles of bromide ions in each solution.
Let's start by calculating the number of moles of bromide ions in each solution using the formula:
moles = concentration (in M) * volume (in L)
a) For 100 mL of 1.0 M BaBr2:
Volume = 100 mL = 100/1000 = 0.1 L
Moles = 1.0 M * 0.1 L = 0.1 moles of bromide ions
b) For 50 mL of 2.0 M NaBr:
Volume = 50 mL = 50/1000 = 0.05 L
Moles = 2.0 M * 0.05 L = 0.1 moles of bromide ions
c) For 20 mL of 2.5 M FeBr3:
Volume = 20 mL = 20/1000 = 0.02 L
Moles = 2.5 M * 0.02 L = 0.05 moles of bromide ions
d) All contain the same number of bromide ions: This option is false because the calculations above show that the solutions contain different numbers of moles of bromide ions.
Therefore, the solution with the greatest number of bromide ions is Option b) 50 mL of 2.0 M NaBr, which contains 0.1 moles of bromide ions.