Dr. Neutron

This page lists questions and answers that were posted by visitors named Dr. Neutron.

Questions

The following questions were asked by visitors named Dr. Neutron.

Answers

The following answers were posted by visitors named Dr. Neutron.

1st: (43.4/31.9988= 1.356 2nd: 133500861/2=.678 3: 44.0098*.678=29.84526482g your answer is 29.8g

first you take (3.16x10^-8)^3)=3.15^-23 thenyou take the 2 atoms and divide it by (3.15x10^-23)= 6.34x10^22 then:you take(6.34x10^22)and divide it by( 6.022x10^23)= 0.105 last you take 19.35 divid by 0.105= and your answer is 184g*mol

1st: 100ppm= 0.1g/L or0.100g/L so 134ppm= 0.134g/L. 2nd:.134g/L divide by molar mass of CaCO3 which is 108.09g/mol. Gram and gram cancel out left you with mole and liter Answer: .00124 mol/L

equation: ln(P2/P1)=(delta H/R)((1/T1)-(1/T1) 1st:T1=78.4+273=354.4K T2= 15+273=286K R=8.3154J/mol delta H=38.6*1000=38600J/mol 2nd: ln(P2/1atm)=((8600J/mol/8.3145J/mol)((1/354.4)-(1/286))= 3.0211 3rd: natural e^3.0211= 20.513 4th: (P2/1atm)=(1/20.513) 5t...

the answer is D C2H6(g)has a standard entropy of 206.5 and C2H4(g) has a standard entropy of 219.5 which are higher than all the other substances.

1st:7= -log( 4.2x10^-7)+log(NaHCO3÷.100) 2nd: 7=6.3767+log(NaHCO3÷.100) 3rd: minus6.3767 on both side which equal .6232 lead to -->.6232= log(NaHCO3÷.100) 4th: 10^.6232=10^log(NaHCO3÷100) 5th: 4.2= (NaHCO3÷100) 6th: cross multiply:4.2x.100=4.2mole now NaH...

1st: equation: H2CO3+HCl<-->H2CO3+Cl 2ndICE For Initial: H2CO3=.42, HCl=.02,H2CO3=..100from?1 For Change:H2CO3=-.02, HCl=-.02,H2CO3=.+.02 For Equilium:H2CO3=.40, HCl=0,H2CO3=.120 2nd:Use Buffer equation: pH= -log(4.2x10-7)+log(.40÷.120)=? 3rd: answer:7pH

1st: equation: H2CO3+NaOH<-->H2O+HCO3 2ndICE For Initial: H2CO3=.100, NaOH=.02,HCO3=..42 fromquestion1 For Change:H2CO3=-.02, NaOH=-.02,HCO3=.+.02 For Equilium:H2CO3=.08, NaOH=0,HCO3=.40 2nd:Use Buffer equation: pH= -log(4.2x10-7)+log(.40÷.08)=? 3rd: answ...

There are a mistake in the calculation. The correct calculation 4.2x.100=.42mole now NaHCO3=.42mole 7th: multiply molar mass: .42x84g/mole=? answer: 35gram not39gram

equation : B4O5(OH)4+2H+3H2O-->4H3BO3 1st convert 12mL to L:12mL/1000mL=.012L and 5mL to L 5mL/1000mL=.005L 2nd Calculate mole of HCL: .800M(mol/L)*.012L=.0060mol 3rd Calculate mole of B4O5(OH)4th:.0060/2=.0030mol 4th Calculate B4O5(OH)concentration: .003...

a).500Mx.012L=.006mole b) ..5*.006=.003moles c) .003mole/.005Liter=.0M d)4*.6^2=.90=Ksp

You use the wrong equation. Use this equation square root (a^3/M) and you will get the answeranswer equal to .110yr ais1 and M is 82kg

For a, it is less than and for b,itis The square root of 1^3/82=.110yr

They're all wrong. Use this equation the Square root of a^3/M)nbwith a=1AU and M=82kg and you will get the answer that you want.