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What is the pH of a solution which has a hydroxide ion OH- concentration of 1.25 x 10-5?
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If 55 ml of a 0.211 m NaHO is diluted to a final volume of 125 ml what is the concentration of naoh in the diluted solution?
Are you sure you mean 0.211 m not 0.211 M Is the concentration molality or molarity? If 55 ml of a 0.211 m NaOH is diluted to a final volume of 125 ml what is the concentration of NaOH in the diluted solution?NaOH not NaHO the compound is Sodium hydroxide Molar mass of NaOH = 23 + 16+ 1 = 40 0.211 moles of NaOH = 0.211 * 40 = 8.44 grams of NaOH per liter of solution 55 ml = 0.055 Liter 8.44 grams of NaOH per liter of solution * 0.055 liter = 0.4642 g of NaOH Moles of NaOH = 8.44 ÷ 40 = 0.211 moles 0.211 moles of NaOH in 125 ml = 0.211 moles ÷ 0.125 L = 1.688 m
How many grams of naoh are needed to prepare 500ml of 125M naoh?
I think you may have missed a decimal point somewhere. 125M of NaOH would be a solution of sodium hydroxide containing 125 moles per litre. One mole of a compound is the same number of grams as the molecular weight of the molecule. Sodium hydroxide has a molecular weight of 40 ( sodium 23, oxygen 16, and hydrogen 1), so a one molar solution would have forty grams of NaOH per litre. 500ml of a 1M solution would contain 20g. 500ml of a 125M solution would need 2 500g. 1L of a 125M solution would need 5 000g of sodium hydroxide in the litre. The maximum solubility for NaOH in water at 20 degrees is 1110g per litre, so if you tried to dissolve 5 000g in a litre you would be left with 3 890g undissolved. A 1.25M solution would have 1.25 times 40g per litre, which is 50g per litre. 500ml of this solution would have half this amount of NaOH, or 25g.
What is the molarity of acetone in water?
The question, as worded, is a little ambiguous. Rather, the question you should be asking is “What is the molarity of a 125 ml aqueous solution containing 10.0g of acetone?” Acetone is roughly 58 grams per mole. Therefore, a 125 mil solution with 10 g of acetone would contain roughly 0.17 moles, and the molarity would be roughly 1.4See the Related Questions for more information about how to calculate the molarity of a solution
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The common factors of 105 and 125 are: 1 and 5
If 55 ml of a 0.211 m NaHO is diluted to a final volume of 125 ml what is the concentration of naoh in the diluted solution?
Are you sure you mean 0.211 m not 0.211 M Is the concentration molality or molarity? If 55 ml of a 0.211 m NaOH is diluted to a final volume of 125 ml what is the concentration of NaOH in the diluted solution?NaOH not NaHO the compound is Sodium hydroxide Molar mass of NaOH = 23 + 16+ 1 = 40 0.211 moles of NaOH = 0.211 * 40 = 8.44 grams of NaOH per liter of solution 55 ml = 0.055 Liter 8.44 grams of NaOH per liter of solution * 0.055 liter = 0.4642 g of NaOH Moles of NaOH = 8.44 ÷ 40 = 0.211 moles 0.211 moles of NaOH in 125 ml = 0.211 moles ÷ 0.125 L = 1.688 m
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What is the solution of -5x25?
-125
What is 12.5 percent of 84?
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What are the factors of 105 using exponents?
(1, 105) (3, 35) (5, 21) (7, 15)
Numbers greater than 105 but less than 125?
106
What is the GCF of 125 175?
The GCF of 125, 200 is 25.
Which is a prime number 247 111 125 152 83 or 105?
83
