Wednesday

April 23, 2014

April 23, 2014

Total # Posts: 803

**Chemistry**

Final Volume=Initial Volume + Volume of water delivered Use the density of water and solve for volume: Density=mass/volume So, Volume=mass/density Volume of water delivered=2.67g/1.00g/mL Grams cancel out and you are left with volume (mL) Volume of water delivered=2.67mL Final...

**Bio**

During interphase, the DNA content is doubled. After cytokinesis, the DNA content is cut in half for the daughter cells. Therefore, I would rule out cells 1, 4, and 7. After cytokinesis, the cells should have half the DNA content as the mother cell, which leaves cells 2 and 5....

**Science**

A. Halfway between the highest point and the lowest point of a jump

**Chemistry**

Partial pressure=(moles of Ar/total moles of gas mixture)*total pressure Partial pressure of Ar=(0.48mol/1.42mol)*1.68 atm ****Answer contains three significant figures.

**Chemistry Need Help!**

The limiting reagent is Naphthalene, which means that the moles of NitroNapthalene will equal the moles of Napthalene. Using your calculations, but with two extra-significant figures, moles of Napthalene=moles of NitroNapthalene. 0.02341 moles of Napthalene=0.02341 moles of Ni...

**Chemistry**

Steve gave you the combined gas law, but it seems that you were confused. The gas law is as followed: P1V1/T1=P2V2/T2 Where P1=75.0 cmHg V1=500.0 mL T1=273K +(-185ºC)=88K P2=55.0 cmHg V2=221.0mL and T2=? Solve for T2: T2=[T1*(P2V2)]/P1V1 T2=[88K*(55.0 cmHg*221.0mL)/(75.0 ...

**Chemistry - DrBob222**

I do not know who provided you the names, but 1-chlorohexane is not valid: it should just be Chlorohexane. Chlorohexane and dodecane are the two primary products from the termination step in free radical halogenation, as well as HCl, therefore you can rule those two answers ou...

**Chemistry urgent**

I would have to say hex-1-ene.

**Chemistry**

USe the combined arrehiuns equation and solve for Ea: ln(k1/k2 )=Ea/R(1/T2−1/T1) Where k1=1.75 x 10^-4 T1=723K k2=2.19 x 10^-3 T2=769K R= 8.314 J/mol*K And Ea=? ln(1.75 x 10^-4/2.19 x 10^-3 )=Ea/8.314 J/mol*K(1/769K−1/723K) -2.5269=Ea/8.314 J/mol*K(-8.2736 x 10^-5)...

**physics**

Gave the wrong units, which is J/s; I apologize, I was moving too fast, but the set up is correct.

**physics**

(J) momentum=mass* change in velocity J=3kg*(10m/s-0m/s) since the object started at rest, the initial velocity =0m/s. J=30 J

**physics**

F=ma F=m(âˆ†v/t) Solving for timt (t): t=(mâˆ†v)/F Where t=? m=20.0kg âˆ†v=8m/s-3m/s=5m/s F=100N Solve for t: t=(20kg*5m/s)/100N t=1s

**Anatomy & Physiology**

You are correct: it is B Glycogen.

**chemistry**

I think he meant how many CO2 molecules are formed from a combustion reaction using methanol (CH3OH) as the hydrocarbon. CH3OH + O2 ----> CO2 + H2O Balance the reaction: 2 CH3OH + 3 O2 ----> 2 CO2 + 4 H2O The reaction shows that 1 mole of CH3OH produces 1 mole of CO2. So...

**Biology**

If the primers are not complimentary then DNA replication will not take place. IF they are complimentary then DNA replication will take place.

**Physics**

Sum of Forces in the Y-direction is the following: T*Sin(theta)=mg Solve for theta theta=Sin^-1(mg/T) Theta=Sin^-1[(0.160kg*9.8m/s^2)/14.5N] Theta=6.2Âº

**physics**

12 bricks/min*(1min/60s)=0.2bricks/s m=(3.75kg/brick) F=mg=(3.75kg/brick)*9.8m/s^2=36.75N/brick d=3.3m Since Power=Work/t(s)=F*d/t Power=(36.75N/brick)*(0.2bricks/s)*(3.3m)=24.3 W ***Bricks cancel leaving (N*m)/s 24.3 W*(1hp/745.70 W)=0.0325 hp

**Physics I**

Graham is correct; I just do not like the way that it is explained in some examples solved. I saw this earlier, but left it alone, and went and did something else. The resultant force points up and you have the force of gravity point down subtracting from the weight. So the re...

**biochemistry**

Some of the texts are showing funny. So, I do not want to tackle this problem. Best

**Physics 210**

Vy(f)^2=Vy(i)^^+2gd Where Vy(f)=0m/s Vy(i)=? g=9.8m/s^2 d=3.10 Solve for Vy(i) 0=Vy(i)^2+2(-9.8)(3.10m) Vy(i)^2=60.8m Vy(i)=sqrt*[60.8m] Vy(i)=7.79m/s This is the velocity in the y-direction: V(i)*Sin35º=7.79m/s Solve for V(i): V(i)=7.79m/s/Sin35º V(i)=13.6m/s

**Physics 210**

Vy(f)^2=Vy(i)^^+2gd Where Vy(f)=0m/s Vy(i)=? g=9.8m/s^2 d=3.10 Solve for Vy(i) 0=Vy(i)^2+2(-9.8)(3.10m) Vy(i)^2=60.8m Vy(i)=sqrt*[60.8m] Vy(i)=7.79m/s This is the velocity in the y-direction: V(i)*Sin35º=7.79m/s Solve for V(i): V(i)=7.79m/s/Sin35º V(i)=13.6m/s

**physics**

18m/s*Cos53º=Vx

**physics**

Relationships or information needed: Density of water=1g/cm^3 3,600s=1hr 1L=1 m^3=10^3 cm^3 1kg=10^3g Solve for seconds in 17hr: 17hr*(3600s/1hr)=seconds 6.20L*(1 m^3/1L)*(10^3 cm^3/1m^3)=6.20 x 10^3 cm^3 Solve for mass of water: Density=mass/volume So, density*volume=mass (1g...

**Physics**

I solved a problem exactly like this the other day, but I made a math mistake; I should have checked my math. Equation 1.) R1+R2=783.4 Equation 2.) R1*R2/(R1+R2)=Req=783.4 Substitute equation 1 into 2: R1=783.4-R2 and R1*R2/(R1+R2)=Req=171.3 (783.4-R2)*R2/783.4-R2+R2=171.3 R2^...

**physics**

Equation 1.) R1+R2=783.4 Equation 2.) R1*R2/(R1+R2)=Req=783.4 Substitute equation 1 into 2: R1=783.4-R2 and R1*R2/(R1+R2)=Req=171.3 (783.4-R2)*R2/783.4-R2+R2=171.3 R2^2-783.4R2=-1.342 x 10^5 R2^2 -783.4R2+1.342 x 10^5=0 (R2 -3.663 x 10^2)^2 R2=3.663 x 10^2 Ω R1 + R2=783.4...

**physics**

Yes, but the for A the speed is 5.5m/s and 0 for the velocity. I must have been tired. I just didn' t read the question carefully.

**physics=> Steve**

Steve you may be correct, but both questions can be worded a little bit better. For A: I wasn't sure if d=7.4m or d=2(7.4)=14.8m. If so, then r=14.8m/2.7s=5.5m/s. For B: I wasn't sure if they wanted to know the average velocity for the trip up, or the trip up and back ...

**physics**

Part A d=r*t distance=rate*time 7.4m=r*2.7s 7.4m/2.7s=r r=2.7m/s Part B Use the following equation: Vf^2=Vi^2 +2 ad Where: Vf=0 a=g=9.8m/s Vi=? d=7.4m 0=Vi^2 +2(-9.8m/s^2)*(7.4m) 0=Vi^2-145m^2/s^2 145m^2/s^2=Vi^2 sqrt*(145)=Vi Vi=12.0 m/s AverageV=(Vf+Vi)/2=(0+12.0m/s)/2=6m/s

**chemistry**

Yes, a typo.

**chemistry**

The reaction is as followed: 2KClO3 ---(MnO2)---> 2KCl + 302 MnO2 is a catalyst 2 moles of KCl produces 3 moles of O2 0.72 g of KCl*(1 mole/74.5513 g)= moles of KCl moles of KCl*(3 moles of O2/1 mole of KCl)= moles of O2 moles of O2*(31.999g/mole)= mass of O2 (Mass of O2/0....

**Chemistry**

The reaction is as followed: B + H2O ------------> BH + OH^- Kb=[BH][OH^-]/[B] At equilibrium, BH=OH^- B=0.250M-OH^- pH+pOH=14 14-pH=pOH 14-12.72=1.28 pOH=-log[OH^-] OH^-=10^(-pOH) OH^-=10^(-1.28) OH^-=5.25 x 10^-2 M BH=5.25 x 10 ^-2 M B at Eq=B-OH^-=0.25M-5.25 x 10^-2=0.19...

**Science Physics**

The first post probably wasn't that clear F1=force of 2.8m ramp F2=force of 0.9m ramp Work=F*d So, Work=F1*2.8m Work=F2*0.9m F1*2.8m=F2*0.9m F1/F2=0.9m/2.8m (0.9m/2.8m)*100=32.4% F1is 32.4% of F2

**Science Physics**

Work=F*d So, Work=F1*2.8m Work=F2*0.9m F2*2.8m=F1*0.9m F2/F1=0.9m/2.8m (0.9m/2.8m)*100=32.4% F2 is 32.4% of F1

**Science/Biology**

C.) Seems to be the most plausible answer.

**chemistry**

You need to create an identity to solve for the mass of fluorine: 1.65 kg of Mg=2.58 kg of F and 1.38 kg of Mg=x kg of F So, x kg of F/1.38kg of Mg=2.58 kg of F/1.65 kg of Mg Solve for x kg of F, x kg of F=(2.58 kg of F/1.65 kg of Mg)*1.38kg of Mg x kg of F=2.157Kg of F 2.157 ...

**chemistry**

You need to create an identity to solve for the mass of fluorine: 1.65 kg of Mg=2.58 kg of F and 1.38 kg of Mg=x kg of F So, x kg of F/1.38kg of Mg=2.58 kg of F/1.65 kg of Mg Solve for x kg of F, x kg of F=(2.58 kg of F/1.65 kg of Mg)*1.38kg of Mg x kg of F=2.157Kg of F 2.157 ...

**Science**

It has a function that keeps the cell alive, just like an organ helps to keep a person alive.

**physics**

D) velocity decreasing, acceleration zero Velocity can not change if there is no acceleration.

**phsics**

Use the following formula and solve for Vf Vf=Vi+at Where Vf=? Vi=60m/s a=-10m/s^2 t=12s **Since the acceleration is in the opposite direction of the velocity, acceleration is negative. Vf=60m/s+[(-10m/s^2)(12s)] Vf=-60m/s Velocity is in the opposite direction, then velocity i...

**physics**

No, B. Its an average

**Physics**

The statement that the velocity and acceleration MUST be in the SAME DIRECTION is not correct. A particle can have an initial velocity in one direction and have an acceleration in another. Over time, the particle's velocity will not be in the opposite direction of the acce...

**Physics**

If a particle's speed increase over a time period this means that it has acceleration, therefore you can rule out answer choices D and E. If a particle's speed increases, its' acceleration can not be negative, and therefore must be positive. Therefore, you can rule...

**Physics**

D=RT Total distance of the the second car is 60miles +x Total distance of the first car is x Two equations: 1.) 60miles+x=60mphT 2.) x=(45mph)T Substitute 2 into 1 gives the following: 60miles + 45mphT=60mphT Solve for T: 60miles=15mphT 60miles/15mph=15mph/15mphT 4hr=T Answer ...

**Physics**

D=RT Total distance of the the second car is 60miles +x Total distance of the first car is x Two equations: 1.) 60miles+x=60mphT 2.) x=(45mph)T Substitute 2 into 1 gives the following: 60miles + 45mphT=60mphT Solve for T: 60miles=15mphT 60miles/15mph=15mph/15mphT 4hr=T Answer ...

**Physics**

Use the following formula: Vf=Vi^+gt Where t=? Vi=(19m/s*Sin33Âº) Vf=0 g=-9.8m/s^2 Solve for t, Vi=(19m/s*Sin33Âº)+(-9.8m/s^2)t 0=10.35m/s-9.8m/s^2*t -10.35m/s=-9.8m/s^2*t (-10.35m/s)/9.8m/s^2=-9.8m/s^2/-9.8m/s^2*t t=1.06 for the ball to reach the top of ...

**Physics-Typo**

That is the initial velocity in the y-direction. Use the following formula: Vf^2=Vi^2+2gd Where d=? Vi=(27m/s*Sin71º) Vf=0 g=-9.8m/s^2 Solve for d, 0=(27m/s*Sin 71º)^2+2(-9.8m/s^2)d 0=651.7m/s-19.6m/s^2*d (-651.7m/s/-19.6m/s^2)=d d=33.25m=33.3m

**Physics**

That is the initial velocity in the y-direction. Use the following formula: Vf^2=Vi^2+2gd Where d=? Vi=(27m/s*Sin71º)+2(-9.8m/s^2)d Vf=0 g=-9.8m/s^2 Solve for d, 0=(27m/s*Sin 71º)^2+2(-9.8m/s^2)d 0=651.7m/s-19.6m/s^2*d (-651.7m/s/-19.6m/s^2)=d d=33.25m=33.3m

**Physics**

That is the initial velocity in the y-direction. Use the following formula: Vf^2=Vi^2+2gd Where d=? Vi=(27m/s*Sin71º)+2(-9.8m/s^2)d Vf=0 g=-9.8m/s^2 Solve for d, 0=(27m/s*Sin 71º)^2+2(-9.8m/s^2)d 0=651.7m/s-19.6m/s^2*d (-651.7m/s/-19.6m/s^2)=d d=33.25m=33.3m

**Physics**

t's are on both sides of the equation, so you can cancel out one t on each side, and the equation becomes the following: 1/2(2.54m/s^2)t=(36m/s) solve for t 1.27t=36 t=36/1.27 t=28.3s

**Physics**

t's are on both sides of the equation, so you can cancel out one t on each side, and the equation becomes the following: 1/2(2.54m/s^2)t=(36m/s) solve for t 1.27t=36 t=36/1.27 t=28.3s

**Physics**

d=v*t and d=1/2at^2 For speeder, d=v*t=(36m/s)*t For officer, d=1/2at^2=1/2(2.54m/s^2)t^2 set equations equal to each other: 1/2(2.54m/s^2)t^2=(36m/s)*t t's are on both sides of the equation, so you can cancel out one t on each side, and the equation becomes the following:...

**Science**

I was thinking of typing in an answer that outlined steps for this question to determine the information needed in the question, but the previous question led me to believe that the two are connected and that some type of information is needed.

**Science**

You need the mass of the bat and the volume. Or you need information that will allow you to figure the above information out. Without it, no one can help you.

**Science**

You need the mass of the bat and the length in some-type of units. Or you need information that will allow you to figure the above information out. Without it, no one can help you.

**AP Chemistry**

Find the number of moles of Fe2O3 in 10.5g of Fe2O3: 10.5g*(1 mole/159.6882 g)=moles of Fe2O3 Convert moles of Fe2O3 into moles of Fe: Moles of Fe2O3*(2 moles of Fe/1 mole of Fe2O3)=moles of Fe How many g of Fe are in the moles of Fe calculated above? Moles of Fe*(55.845g/1mol...

**physics**

***Convert initial speed to m/s a=-6.5m/s^2 Vf=0 d=? ***Vi=97km/h*(10^3m/1km)*(1h/3600s)=27m/s Use the following equation and solve for d: Vf^2=Vi^2+2ad 0^2=(27m/s)^2+2(-6.5m/s^2)d 729=13d Solving for d, 729/13=d d=56m

**Biology (Please Help: Urgent)**

aa=36% or 0.36 <=This the frequency of the aa genotype let aa=p^2 0.36=p^2 p=sqrt*[0.36] p=0.6 <=This is the frequency of the a allele p+q=1 So, 1-p=q 1-0.6=0.4=q q=0.4 <=This is the frequency of the A allele p^2+ 2pq + q^2=1 (0.6)^2+2(0.6)(0.4)+(0.4)^2=1 0.36+0.48+0....

**science**

I do not know how my answer earlier was reposted, but I think I should have went through the math. There is agreement that the velocity will be 1m/s for the first question. However, there is disagreement concerning the velocity for the second question. Let M1=300 kg V1i=0 V1f=...

**science**

This is a momentum problem and momentum is conserved: M1V+M2V=M3V Let M1=75 kg M2=25 kg M3=300 kg V1=3 m/s V2= 3 m/s V3=? (3m/s)*(75 kg +25 kg)/300 kg=V3 v=1 m/s Second problem: Answer Should be the same; momentum is conserved. But I am not 100% sure.

**science**

This is a momentum problem and momentum is conserved: M1V+M2V=M3V Let M1=75 kg M2=25 kg M3=300 kg V1=3 m/s V2= 3 m/s V3=? (3m/s)*(75 kg +25 kg)/300 kg=V3 v=1 m/s Second problem: Answer Should be the same; momentum is conserved. But I am not 100% sure.

**Maths/Physics**

I think I know what you are looking for, so starting off where Damon left off: Max acceleration occurs when the velocity=acceleration. Vi=0m/s^2 Vf=5m/s and a=5m/s^2 d=? Solve for d. Vf^2=Vi^2+2ad (5m/s)^2=0+2(5m/s^2)d 25=10d 25/10=d d=2.5m The sprinters will not keep accelera...

**Physics URGENT HELP**

I think I know what you were looking for. Starting off where Damon left off, Max acceleration occurs when the velocity=acceleration. Vi=0m/s^2 Vf=5m/s and a=5m/s^2 d=? Solve for d. Vf^2=Vi^2+2ad (5m/s)^2=0+2(5m/s^2)d 25=10d 25/10=d d=2.5m The runner establishes max acceleratio...

**Physics URGENT HELP**

I think I know what you were looking for. Starting off where Damon left off, Max acceleration occurs when the velocity=acceleration. Vi=0m/s^2 Vf=5m/s and a=5m/s^2 d=? Solve for d. Vf^2=Vi^2+2ad (5m/s)^2=0+2(5m/s^2)d 25=10d 25/10=d d=2.5m The runner establishes max acceleratio...

**Chemistry**

Dr. Bob222 gave you the setup to calculate the grams of CaCl2. But I will go into a little bit more detail and maybe this will help: Molarity=Moles/Volume (L) The question tells you that you need need a total of 2L, but you don't know how many moles are in a 3.5M solution ...

**Chemistry**

You will only see evidence of it if you use enough base to cause the di or polyprotic acid to donate the 2nd or subsequent protons. 10mL of a 0.1 M solution of H2SO4 is a a di protic acid. If I only add 10mL of a 0.1 M solution NaOH then I will not see signs that it is a dipro...

**chemistry**

X = propanol==> CH3-CHOH-CH3

**chemistry**

I'll try it again CH3-CH=CH2 <=Y=propene CH3-CO-CH3 <=Z=propanone

**chemistry**

CH3-CH=CH2=Y=propene CH3-CO-CH3=Z=propanone

**--->CHEMISTRY<--- HELPP**

No and yes. .001 to the right of 5.623 x 10^2. So the answer should be 5.520 x 10^3

**chemistry=>Dr.Bob222**

I was wondering why the individual ask the question when it seemed kind of straight forward. Sometimes I want to answer a question, but I am not sure what they are asking, or I'm not sure concerning the concept if I haven't done a particular type of problem in a while....

**chemistry=>Dr.Bob222**

Okay!!!!!!!

**chemistry**

If the volume is split, then the amount of moles will be split as well. However, the mole to volume ratio will not change in a homogenous mixture. So, the molarity can be solved using MV=MV M=2.5M*(21.8mL/12.5mL) Answer contains 2 significant figures.

**Dr. Bobb222**

You were more careful then I was as I child. Sometimes I would get careless, or I would just move to fast and not check my work; I do that a lot with my typing. I didn't start applying myself in school until my second year in college. Once I start applying myself, I starte...

**Dr. Bobb222**

Okay, the answer choices are: 375 torr 119 torr 195 torr 76.3 torr When I performed the calculations, I also calculated 196 torr after rounding up, so I wasn't sure. I don't know if you remember ever taking a multiple choice exam that required using math as a kid and w...

**Dr. Bobb222**

Okay, the answer choices were The answer choices are: 375 torr 119 torr 195 torr 76.3 torr When I performed the calculations, I also calculated 196, so i wasn't sure. I don't know if you remember ever taking a multiple choice that required using math as a kid and when ...

**Dr. Bobb222**

When you get a chance: If the pure liquid vapor pressure of benzene is 155 torr and the pure liquid vapor pressure of chloroform is 333 torr, then what is the total vapor pressure at 30°C of a solution of 9.26 g of benzene (C6H6) and 4.26 g of chloroform (CHCl3)? I get 195...

**chemistry**

One other thing: when you do this, make sure to shake it vigorously for at least 45 seconds if doing it by hand, but I would prefer that you use a vortexer for at least 45 seconds to make sure that it is mixed properly. I know when I needed to make dilutions or needed a dilute...

**chemistry**

Your chemistry teacher is correct. The reaction is exothermic, so adding water to acid will cause it to splash around and it may get on you, or someone else in the lab, which isn't good lab safety protocol. The proper way is to add the acid to water, and yes, you should ad...

**chemistry**

You can add 6 mL to 94 mL, but the total volume may or may not be exact. The method that Dr. Bob222 recommended is the correct method. If you were to add 6mL of the acid to volumetric flask and then add 94 mL of DI water to it, then the bottom of the meniscus may be above or b...

**biology**

Where are the results after four generations? Repost later with the information and maybe someone will help you solve this problem.

**biology**

c=0.16 C+c=1, so 1-c=C C=1-0.16=0.84 (C+c)^2=C^2+2Cc+c^2=1 (0.84)^2+[2(0.84)(0.16)]+(0.16)^2=1 Cc=Heterozygous dominant Cc=2(0.84)(0.16)=0.27 CC=Homozygous dominant CC=(0.84)^2=0.48 Frequency of CC + the frequency of Cc multiplied by 100 =percentage dominant for the trait (0.2...

**biology**

b=0.4 B+b=1, so 1-b=B B=1-0.4=0.6 (B+b)^2=B^2+2Bb+b^2=1 (0.6)^2+[2(0.4)(0.6)]+(0.4)^2=1 Bb=heterozygous Bb=2(0.4)(0.6)=0.48 But if you need to know, BB=0.36 bb=0.16

**chemistry**

As long as the container is sealed, conservation of mass is conserved (i.e., 1,000 g is before the reaction and 1,000g after the reaction.)

**Chemistry==> Dr. Bob222**

Should say over the past month.

**Chemistry**

Do you know how many times that I have done that? If you check some of my posts over the years, you will see me saying the same thing.

**Chemistry**

I used the two equations in my original post and set them to each other and solved for E° ∆G°=∆H°-T∆S°=−nFE° [∆H°-T∆S°]/-nF=E° I apologize for not showing that initially.

**Chemistry**

You will need the following equations: ∆G°= ∆H°-T∆S° ΔG°=−nFE° Where ∆H°=-76.8kJ ∆S°=-225J/K n=3 T=273+25°C=298K F=9.65 x 10^4 J/ V *mol Solve for E°: −nFE°= ∆H°-T∆S° E&d...

**Analytical chemistry**

19.51 x 10^-3 L*0.1060M=2.07 x 10^-3 moles of HCl was used, so 2.07 moles of H2BO3- reacted with HCl., which means that 2.07 x 10^-3 moles of NH4^+ was collected. Since we know that 2.07 x 10^-3 moles of N is present, we need to find out how much C2H7N5 is present in the sampl...

**Analytical chemistry**

I thought it was an easier way to do it, but it was super early/late and I couldn't reason it at the time.

**Analytical chemistry**

I don't remember ever doing a problem like this, so I am not sure about my answer, so hopefully someone comes along and fixes my mistakes if I made any. 0.2445 g*(1 mole/106 g of Na2CO3)=2.307 x 10^-3 moles of Na2CO3 The reaction shows 2HClO4 are needed for 1 Na2CO3, so 2*...

**biology multiple choice**

2. C 6.A 7. E 8. D 9. B 11. A 12. C 13. D 14.A 15. C 19. C If I agreed with your answers, I deleted it, but if I didn't I gave you my answer choice, which was most of your work. Try not to do this again because I will ignore it, and the other people who come on this site ...

**chemistry**

HCl is a strong acid, so the pH=-log[Molarity of HCl] Calculate the molarity of HCl Molarity of HCl=[(0.87mL)*(1.979)]/12.96mL] Plug in the numbers and solve.

**Chemistry help**

Forgot a step: pH=-log[H3O+], so 10^-pH=H3O+

**Chemistry help**

NaOH is a strong base, so it will completely dissociate into OH-. pOH+pH=14, so 14-pOH=pH Solve for pOH 2.00g of NaOH*(1 mole of NaOH/39.997 g)= moles of NaOH moles of NaOH/2.00L=molarity of NaOH -log[molarity of NaOH]=pOH 14-pOH=pH

**chemistry**

One other thing, your answer should only have 2 significant figures.

**chemistry**

60% (m/m)=60g of ethanol out of 100g in solution. 60g of ethanol*(1 mole/46.07g)= moles of ethanol Density=mass/volume Solving for volume, Volume=mass/density Total volume=100g total solution/0.8937g/ml Molarity=moles/volume in L Molarity of ethanol solution=moles of ethanol/T...

**chemistry**

Since I don't know what figure you are referring to in your later post, I do not want to try and help you solve that one. Maybe Dr. Bob222 will assist you with that one.

**chemistry**

What Dr. Bob222 did was give you the setups to calculate the molarity. Transmittance is related to absorbance, so you have to calculate the absorbance of the blank and subtract it from your sample to to acquire the correct absorbance value for your sample. A unknown- A blank= ...

**Chemistry- Please Help!!**

K=products/reactants=[Fe3+][3OH]^3

**chemistry**

What fashion?

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