- log(2.7 X 10 -3 M HCl)
= 2.5686
14 - 2.5686
= 11.4314
===========now,
1/10(11.4314)
= 3.7 X 10 - 12 M OH -
=================
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I always start a question like this by making 100 mL. 25% of 100 mL is 25 mL. So 25 mL HCl + 75 mL of whatever you are making it in (hopefully water).If your final volume is not an easy number, just multiply 25% times the final volume to get the volume you need of HCl, then subtract that number from your final volume to get the mL of the other component (water). When you add the two together, you should get your final desired volume. NOTE: I need to retract this answer for acids. The below answer is more correct than mine for HCl. My answer works for pure liquids.ANS#2:Always add acid to water, never add water to the acid when you prepare a dilution. Hydrochloric acid exists only as a solution of gaseous Hydrogen Chloride dissolved in water. Since it is a gas before dilution, you will need to calculate your ratios by weight/volume (w/v). At standard conditions the most concentrated solution of HCl(aq) is 40% (w/v). 38% (w/v) is known as 'concentrated grade'. 27% (w/v) is commonly available. You will be starting almost where you need to be if you are using 27% acid.Initial data:100 ml of 27% (w/v) has 27 grams of HCl dissolved into it. 25% would by the same reasoning, have 25 grams of HCl dissolved. The gram molecular weight of HCl is 1.00794 g/mol + 35.453 grams/mol = 36.461 grams/mol. That means we need to get either 25 grams of HCl or 0.686 moles into the final 100 ml. Calculations:Say you start with 27% (w/v) solution: 100 ml has 27 grams of HCl so (27/100 = 25/x , x = 92.59) 92.59 ml has 25 grams of HCl. Put 92.59 ml of the 27% solution into a container and slowly pour it into a container holding (100 ml - 92.59 ml =) 7.4 ml of distilled water. Et voila, you have 25% (w/v) Hydrochloric Acid. More Fun:You could also bubble the gaseous products resulting from reacting 0.686 moles of NaCl with excess H2SO4 through 100 ml of water and you have the same. 25% (w/v) HCl. (The calculations have been updated to correct the gmw for Hydrogen to 1.00794 g/mol.)
Given: 27 mL of NaOH, 0.45M; 20 mL HCI Need: M of HCI 27 ml NaOH*(1 L NaOH/1000mL NaOH)*(0.45M NaOH/1L NaOH)*(1mole HCI/1 mole NaOH)=0.012 0.012/0.02=0.607 M HCI (or rounded 0.61 M HCI)
mass of proton 1.67262192369(51)×10−27 kg NB Note this in kg (kilograms) which is the normal S.I. unit of mass.
1836 times the mass of an electronSo how do you calculate '1836 times the mass of a proton' ?Mass of proton = 1.673 x 10-27 or (10 to the power of negative 27)Mass of electron = 9.109 x 10-31 or (10 to the power of negative 31)Simply by dividing the largest mass (of protons) by the smallest mass (electrons):(1.67262 . 10-27) / (9.10939 . 10-31) = [1.67262 /9.10939] . [10(-27)-(-31)] =0.1836 . 10+4 = 1836
27 over 10 = 27/10 in fraction
27/10 divided by 9/10same as,27/10 * 10/9= 27/9= 3===
27-8x5+10=?27-40+10=? -13+10=? =-3
27 out of 10 = 270%
10^27 = 1,000,000,000,000,000,000,000,000,000
27/10
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on January 27 1945
Converting 27/10 to mixed number:Divide the numerator by the denominator.The remainder will be the numerator of the fractional part.27/10:= 27 ÷10= 2 r7= 2 7/10therefore: 27/10 = 2 7/10
The GCF for 27 and 10 is 1Factors of 27: 1, 3, 9, 27Factors of 10: 1, 2, 5, 10GCF (27, 10) = 1
27/270 = 1/10 = 10%
27 rounded to the nearest 10 is 30.