6.585 g LiNO3
150 mL x 40.0 g LiNO3/100 mL solution = 60.g of solute
The mass of sugar is 100 g.
You need 6 g potassium iodide.
35.0%
If it's a 4% solution by mass, you want 4.167g of acetic acid (25g/6)
150 mL x 40.0 g LiNO3/100 mL solution = 60.g of solute
50 percent of the dissolved minerals will condense out of solution by crystallization.
40%
This solution contain 5 g NaCl dissolved in water.
9%
The percent of nitrogen in sodium nitrate is 16,47 %.
Bioc 4025? Just a guess. Definition of % Saturation: The amount of a substance that is dissolved in a solution compared to the amount that could be dissolved in it. % w/v Solution: This is in terms of weight (g) per volume (mL). 2% for example is 2 g per 100 mL. gl
25 percent by mass
150 mlLet x be the volume of 20% solution to add. The amount of "stuff" dissolved in the 20% solution plus the amount in the 40% will equal the amount of "stuff" dissolved in the 25% solution.(0.20)(X) + (0.40)(50 ml) = (0.25)(X+50ml)0.2X + 20 = 0.25X + 12.520 - 12.5 = 0.25X - 0.20X7.5 = 0.05XX = 7.5/0.05 = 150 ml
The mass of sugar is 100 g.
Sodium nitrate is NaNO3. The percent nitrate is given by :PCT Nitrate = [ ( 14 + 48 ) / ( 23 + 14 + 48 ) ][ 100 ] = [ 62 /85 ][ 100 ] = 72.94 mass percent
The percent of N in sodium nitrate would be 84.6. This is considered a lot of sodium.