Can someone please check my work PRE-LAB PROBLEMS:1. Calculate logarithms and antilogarithms. You should be able to do the first three withouta calculator.NumberLog1000log Cio00) - 30.01log Clo3)-21 x 10-4-4.02log (o-oa)=-1-69log (300 )=2.473003.22. Complete the following table.[H*][OH]pHРОНA,B,N?.110-1313Acid10-12.0112Basic3102Acid16-1212410lo-7777.Neutal3) pH: 2, PH= -lo9 BI'].2 acid solutionif PH= 7=Neuralit PH >7 - Basicit pH <7 = Acid4) In toble CH'] = 0.1, PH: - log [H'] =-logCOH-] = - log(o.o1)there Fore, pH is I (Soluhon is acidic)POH-14-pH =14-1= 13Now, POH = -log COH-] =13therefore, [oHÝ -10-13CH j= 10-2POH : -10g CoH] = 126H = 10-24) POH : 7, POH: -109[OHI=7COH-]= 1072) COH ] = •ol, POH= -10g COH ]= -1og(0-0l)So POH = 2PH:14- POH = 14-2=12PH 12 (solukon is basic)PH = - log CH] = 12CHJ = 10-12PH = 14- pOH = 1U-7= 7PH = 7 (soluHon is Neuhol)Now pH: -lo9 CH]= 7CH] =107

Hello. Since there are more than one question, the first question shall only be answered in this case. If the answer of the remaining question is required, please specify the question and submit it again. Thanks.The value of log (1000) can be calculated as follows: log (1000) = log(103)= 3 × log(10)= 3 × 1= 3The value of log(0.01) can be determined as follows: log(0.01) = log10-2= -2 × log(10)= -2 × 1= -2Similarly log(10-4) is equal to -4. The value of log (0.02) can be expressed as follows: log (0.02) = log (2 × 10-2)= log2 + log(10-2)= 0.3010 - 2.000= -1.69The value of log (300) can be determined as follows: log (300) = log (3 × 102)= log (3) + 2 log(10)= 0.477 + 2= 2.47The logarithmic value has been given as 3.2. The actual number can be determined as follows: log(x) = 3.2x = 103.2x = 100.2 × 103x = 1.58 × 103