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... = 10^9 nanograms 1 microgram = 1000 nanograms Do you mean molarity (i.e. converting 1 microgram/L to 100nanograms/L?) If so, you would dilute the solution by 10 (so add 1 ...

4 Answers · Science & Mathematics · 20/02/2007

...we want to raise the BP of water from 96.6 C to 100 .0 C, an increase of 3.4 C. delta Tb...i = moles of ions produced by 1 mole of NaCl = 2 Kb = boiling point...

2 Answers · Science & Mathematics · 27/10/2011

... Ni2+: I assume you mean 1 mmol of Ni2+ added to 100 mL of solution. If you do mean 1 mM then you also need a volume added. If 1 mmol then the initial...

1 Answers · Science & Mathematics · 01/04/2015

It depends on the amount of heat it is being exposed to .

1 Answers · Science & Mathematics · 28/08/2010

...and dilute to a total of 1000( add 900 uL of diluent. so for a 1 :30 ...dilute 1 uL to exactly 30uL or dilute 10uL to exactly 300 uL, or 100uL...

1 Answers · Science & Mathematics · 23/06/2008

... per gram (thus approximately 1 calorie per gram); (2) The heat of vaporization of water at 100 °C is 539 calories.../(mol K) at 100 °C ( to raise it to 101°C), or 0.365 ...

5 Answers · Science & Mathematics · 27/01/2009

You take 1 mL of the stock solution, and dilute it to 100mL with distilled water. That's a 1 / 100 dilution.

2 Answers · Science & Mathematics · 23/03/2008

...molar solubility. For the second one it it already in that format, so we need to change only the one of CaF2. The molecular weight of CaF2 is:40+2*19 = 78 So you have 1 .70 mg/ 100 mL = 1 .7 *10^-3 g/100mL = = 1 .7*10^-2 g/L =( 1 .7*10^-2...

1 Answers · Science & Mathematics · 08/03/2007

... Heat) Q=(100grams)(10Celcius)( 1 .00 cal) ------------------------ grams * Celcius... the same as cal ---- g x C = 1000 cal reported to one significant figure

3 Answers · Science & Mathematics · 12/04/2009

The molecular weight of CaC03 ( 100 .02 g/mol) is irrelevant in figuring...something it means you have quantity equal to avogadro's number: 6.022E23. So 1 mole of CaC03 means you have...

2 Answers · Science & Mathematics · 27/09/2011

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