This depends on the mass of NaOH dissolved in 1 L water.
I assume you mean 32.0 grams of NaOH and 450 milliliters of NaOH. Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters ) get moles of NaOH 32.0 grams NaOH (1 mole NaOH/39.998 grams) = 0.800 moles NaOH Molarity = 0.800 moles NaOH/0.450 liters = 1.78 Molar NaOH
The answer is 0,625 moles.
(100ml)(0.125M NaOH) = (500ml)(X Molarity) Molarity = 0.025 M
Molarity = moles of solute/Liters of solution 3.42 M NaOH = 1.3 moles NaOH/Liters NaOH Liters NaOH = 1.3 moles NaOH/3.42 M NaOH = 0.38 Liters
To find the molarity of NaOH, first calculate the moles of HCl: ( \text{moles of HCl} = \text{volume (L)} \times \text{molarity} = 0.020, \text{L} \times 5.0, \text{M} = 0.10, \text{moles} ). In a neutralization reaction, HCl and NaOH react in a 1:1 ratio, so 0.10 moles of NaOH are needed. The molarity of NaOH can be calculated using the formula ( \text{Molarity} = \frac{\text{moles}}{\text{volume (L)}} = \frac{0.10, \text{moles}}{0.100, \text{L}} = 1.0, \text{M} ). Thus, the molarity of the NaOH is 1.0 M.
To determine the molarity of 15 g NaOH in a 100 L solution, first calculate the moles of NaOH using its molar mass (40 g/mol). Then, divide the moles by the volume in liters (100 L) to get the molarity. The molarity of the NaOH solution would be 0.375 M.
To calculate the molarity, you first need to convert the grams of NaOH to moles using the molar mass of NaOH (40 g/mol). Then, you divide the moles of NaOH by the volume of solution in liters (450 ml = 0.45 L) to get the molarity. Molarity = moles of NaOH / volume of solution in liters Moles of NaOH = 95 g / 40 g/mol = 2.375 mol Molarity = 2.375 mol / 0.45 L = 5.28 M
To find the molarity, you need to know the amount in moles of NaOH and the volume in liters. First, convert 10 mL to liters by dividing by 1000 (10 mL = 0.01 L). Then, calculate the number of moles of NaOH using the molarity formula, Molarity = moles/volume. Given that you have 0.05 moles of NaOH and a volume of 0.01 L, the molarity would be 5 M.
I assume you mean 32.0 grams of NaOH and 450 milliliters of NaOH. Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters ) get moles of NaOH 32.0 grams NaOH (1 mole NaOH/39.998 grams) = 0.800 moles NaOH Molarity = 0.800 moles NaOH/0.450 liters = 1.78 Molar NaOH
Yes, NaOH (sodium hydroxide) is typically used in its aqueous form as a caustic alkaline solution.
The molarity of the solution can be calculated by dividing the moles of solute by the volume of solution in liters. In this case, 2 moles of NaOH in 1620 mL (1.62 L) of water gives a molarity of approximately 1.23 M.
Molarity = moles solute/Liters solution get moles NaOH 0.240 grams NaOH (1 mole NaOH/39.998 grams) = 0.0060 moles NaOH ----------------------------------as one to one OH- has this many moles also Molarity = 0.0060 moles OH-/0.225 Liters = 0.0267 M OH- ----------------------- -log(0.0267 M OH-) = 14 - 1.573 = 12.4 pH -------------
There seems to be a misunderstanding, "miles" is not a unit of measurement for NaOH concentration. If you meant molarity instead of miles, you need the molarity of NaOH in order to calculate the moles of NaOH in the given volume which can be converted to miles using the molar mass of NaOH.
The answer is 0,625 moles.
molarity equals moles of solute /volume of solution in litres . moles of NaOH equals 5g/40g = 0.125 and volume of solution will be volume of water + volume of NaOH = 0.5 litre+0.002 l which is nearly 0.5 litre . (volume of NaOH is calculated by its density) so molarity = 0.125mol/0.5litre = 0.25 M
(100ml)(0.125M NaOH) = (500ml)(X Molarity) Molarity = 0.025 M
The reaction between aqueous acetic acid (CH3COOH) and aqueous sodium hydroxide (NaOH) forms water (H2O) and sodium acetate (CH3COONa). The balanced chemical equation is: CH3COOH + NaOH -> H2O + CH3COONa