Recent content by blicker

Homework Statement -------Solution1:-----Solution2:-----Solution3:---------- Solution4:-----Solution5:--------------- Solution6: Exp. 0.200M KI-----2% starch-----0.020M Na2S2O3 -----water----0.050M acetate buffer -----0.40M H2O2...

well thanks for your help anyway.

so would the equation for ants look like this: -(9/15)In(9/15)

yes that's it! but i don't know what numbers go where or how to calculate it. like would i include the sigma symbol and the letters above and below it?

Homework Statement This is the sample population taken along a 5m transect: ants:9 grasshoppers:1 pill bugs:4 spiders:1 Homework Equations What formulas would i use to calculate the species abundance and diversity? The Attempt at a Solution I only know this formula H'=-\Sigma...

Homework Statement cold water: 50g cold water temp: 22.23 C hot water: 50 g hot water temp: 63.2 C final temp: 39.55 C Homework Equations The Attempt at a Solution heat lost by hot water: (4.18)(50)(39.55-63.22)=-4942.85 heat gained by cold water...

no this is a separte question from the first one i asked.

yes that's all he gave us. ill have to ask him about that question. another question i have is how to calculate the heat gained by the water in the calorimeter after adding the aluminum. this is what he gave us: temp of water in calorimeter before adding metal: 22.90 C mass of test tube...

Homework Statement 1. Calculate heat lost by the hot water in the calorimeter. Initial temp: 22.90 C Finial temp: 23.25 C Density: 1.00 g/mL specific heat capacity: 4.18 J/gK Homework Equations qmetal=-(qwater+qmetal) q=m C dT The Attempt at a Solution When i use the second...

Homework Statement Find the mass of water vapor. this is the reaction: NaHCO3+HCl-->NaCl+CO2+H2O V=39.5 L P=18.65 mm Hg T=294 K n=30.5 mol H20 (g) pressure of room= 29.68 inHg=753.87mm Hg Homework Equations The Attempt at a Solution Would i have to use this equation...

yeah i realized my math was wrong on the first reaction. i fixed it. and my teacher said we are suppose to work in that 4.0 mL of HNO3. its suppose to be turned into mols. how would i do that?

i don't know if its right, but i used the answer from the first reaction to calculate the second reaction since all these reactions are connected. 1.08g Cu(NO3)2*(1 mol Cu(NO3)2/187.57g Cu(NO3)2)=0.00576 mol Cu(NO3)2 0.00576 mol Cu(NO3)2*(1 mol Cu(OH)2/1 mol Cu(NO3)2)=0.00576 mol Cu(OH)2...

would i use the answer from the first reaction to calculate the second reaction?

is there more i need to do for the first reaction? and would i do the second reaction the same way?

sorry i didnt mean to stick that g in there. what should i have done differently besides rounding?