how many moles of Ba(OH)2 are present in 295mL of 0.400 M Ba(OH)2?
Molarity = Moles/Liters
therefore
Moles = Molarity x Liters
Convert 295mL to L = 0.295L
Moles = 0.400M x 0.295L
= 0.118 mol Ba(OH)2
To calculate the number of moles of Ba(OH)2, we can use the equation:
moles = concentration (M) x volume (L)
Given:
Concentration (M) = 0.400 M
Volume (L) = 295 mL = 0.295 L
Now, we can substitute the values into the equation:
moles = 0.400 M x 0.295 L
moles = 0.118 mol
Therefore, there are 0.118 moles of Ba(OH)2 present in 295 mL of 0.400 M Ba(OH)2.
To find the number of moles of Ba(OH)2 present in a given volume of a concentration, you can use the formula:
moles = concentration × volume
In this case, the given concentration is 0.400 M (moles per liter) and the volume is 295 mL. However, the volume needs to be converted to liters before using it in the formula.
1 liter is equal to 1000 milliliters, so to convert milliliters to liters, divide the volume by 1000:
295 mL ÷ 1000 = 0.295 liters
Now, you can substitute the values into the formula:
moles = 0.400 M × 0.295 L
moles = 0.118 mol
Therefore, there are 0.118 moles of Ba(OH)2 present in 295 mL of 0.400 M Ba(OH)2.