if log 4 = 0.6020, find the value of log 80. - maths q&a
log 80 can be written as; log (8 ✕ 10) log (4 ✕ 2 ✕ 10) here we can use the formula log (a ✕ b) = log a + log b log 4 + log 2 + log 10  (1) log10 10 = 1 log 4 = 0.6020 log 2 = ½ log 4 = ½ ✕ 0.6020 = 0.3010 substituting the values in equation (1) we get the answer 1.9030
so the server listens on 8080... [readmore]typed answera) δg° = -rtlnksp δg° = -(8.314 jmol-k)(298k)ln(4.6×10-2 )= 7628.72 j = 7.63 kjb) ∆g° = nfe°e° = ∆g°-nfe° = 7.63-(2)(96,000)(11000)e° =-0.0397 v class intervals
frequency
(fi)
xi
di = xi - a
fi di


0 - 20
17
10
-40
-680


20 - 40
28
30
-20
-560


40 - 60
32
50 = a
0
0


60 - 80
p
70
20
20p


80 - 100
19
90
40
760


total
Î£
fi = 96 + p


Î£ fi di = 20p -
480


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/5130830/2013_07_08_13_01_36/mathmlequation8813644107607516415%20png) given = let, logx = z⇒ so,where c is the integrating constant. answer: `1.8706 xx 10^(11) " erg", 4477 cal` 480 >> get(groot,'defaultaxescolororder')
 ans =
 0 0.4470 0.7410
 0.8500 0.3250 0.0980
 0.9290 0.6940 0.1250
 0.4940 0.1840 0.5560
 0.4660 0.6740 0.1880
 0.3010 0.7450 0.9330
 0.6350 0.0780 0.1840
 >> 

the letter monikers for colors are the pure colors, 'b' --> [0 0 1], etc., ... 0 answer: `1.00` a) 3ag + 4hn03 > 3agno3 + no + 2h2o

b) mno2 + 4hcl > 2h2o + mncl2 + cl2