To determine the amount of acid needed to neutralize the base, we can use the formula M1V1 = M2V2, where M1 is the concentration of the acid, V1 is the volume of the acid, M2 is the concentration of the base, and V2 is the volume of the base. Plugging in the values, we get (0.45)(V1) = (1.00)(25.0). Solving for V1, we find that V1 = 55.6 ml of 0.45M HCl is needed to neutralize 25.0 ml of 1.00M KOH.
The quantity of rubbing alcohol present at any moment ina bottle labeled "50-milliliters", expressed in milliliters, is(50) - (number of milliliters previously used for rubbing)
210 ml 1 centiliter = 10 milliliters 1 milliliter = 0.1 centiliter
7 milliliters is equal to approximately 0.24 ounces.
To neutralize HCl with NaOH, the mole ratio is 1:1. So, the moles of HCl are 0.200 M x 0.020 L = 0.004 moles. Since NaOH and HCl react in a 1:1 ratio, we need 0.004 moles of NaOH. Using the molarity formula, we find that we need 0.010 L or 10.00 mL of 0.400 M NaOH.
To convert liters to milliliters, you multiply by 1000. Therefore, a bottle of shampoo with a capacity of 0.43 L contains 430 milliliters of shampoo.
There are 0.045 milliliters in 0.045 milliliters.
100cm = 1m 1cm = 1/100m 216cm = 216/100m = 2.16m
one meter =3.280ft There in 100m we have 3.280*100m =328ft
5.9 milliliters is equal to 5.9 milliliters.
10,000
6000040 milliliters
0.04 milliliters
750
9.7
155 milliliters
I guess you mean how many in a litre - 1000
A concentration of 110 M or 106 M doesn't exist.