0.5 Moles
If you have a 0.25 M solution, you have 0.25 mol/dm3, or 0.25 moles in 1 L (0.25 mol/L)
If you have 2 L of solution, you have 2 L x 0.25 mol/L = 0.5 mol
The L's cancel out, and you're left with moles.
2005 Lincoln cents are .975 Zinc & .025 Copper
A US 1 cent piece prior to 1982 was made of copper and had a mass of about 3.1g. This is about 0.049 moles of copper, or around 2.9 x 1022 atoms. More recent pennies are mostly zinc (97.5%) and have a mass of about 2.5g. This works out to around 2.2 x 1022 zinc atoms and 5.9 x 1020 copper atoms, so overall around 2.3 x 1022 atoms.
This volume is 6,197 399 5 at 25 0C.
1.25 Solution Method: 1. Convert 2.5% to decimal: 2.5 / 100 = .025 2 Compute answer: 50 * .025 = 1.25
0.25 mile - 1320 feet .025 mile - 132 feet
.025 percent Solution Method: 250 ppm = the ratio 250:1,000,000 = the fraction 250/1,000,000 = the decimal value 0.00025 = the percentage .025 (i.e. 0.00025 * 100)
025 is the same as 25 which is greater than 5
.025 meters
.025 miligram
The only information you've given is you want to know moles and how many Liters there are. To calculate, you need to know what you are preparing(i.e., N2, AgNO3, etc.), and it's Molarity(M). Here's an example: How many moles of silver nitrate are needed to prepare 250mL of standard 0.100M silver nitrate solution? Note: M = moles of solute(stuff) / volume of solution(L) = # moles / L So, ?/250mL=.1M First we use dimensional analysis to convert mL to L. 250mL x 10-3 L / 1 mL = .25L Since we know M=moles/L, we can take the Molarity and put it in moles/L form. .10M = .10 mol/L Finally, .10 mol/L x .25L = .025 mol AgNO3 We can make a general assumption for your question and say the M is standard .100M. .10M = .10 mol/L .10 mol/L x 48L = 4.8 moles
It is: .025 times 80 = 2
I assume you mean 5 and 025 (but I'll answer for.025 also). 5 milligrams is larger than 0.025mg. However 5mg is smaller than 025 milligrams.
.025 is thicker than .012
0.0001