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Dr. Neutron
Answers (14)
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.
For a, it is less than and for b,itis The square root of 1^3/82=.110yr
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
a).500Mx.012L=.006mole b) ..5*.006=.003moles c) .003mole/.005Liter=.0M d)4*.6^2=.90=Ksp
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:
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
1st: equation: H2CO3+NaOHH2O+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:
1st: equation: H2CO3+HClH2CO3+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: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
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.
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)
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
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: (43.4/31.9988= 1.356 2nd: 133500861/2=.678 3: 44.0098*.678=29.84526482g your answer is 29.8g
