What is 10% of two(2) gallons?
354060
0.2 gal. or 25.6 fluid oz.
1003456
Depends on the type of gallons as below:\n53 Gallons(British) = 0.241 cubic metres\n53 Gallons(US, dry) = 0.233 cubic metres\n53 Gallons(US, liquid) = 0.2 cubic metres
723851
Dis you mean 2.4 gal?\nif so\n2.4gal*3.78l/gal=9.072l\n9.072l*.692kg/l=6.28kg
757107
10.7%\n\n(1030 - 920) / 1030 = 0.107
1120275
8 to 10 glasses of water a day. Roughly 1/2 gallon
1052913
2/3 of a Tablespoon, or 2 teaspoons
567096
2/3 of 10 miles\n= (2/3)*10\n= 20/3\n= 6 2/3 miles
558772
the correct answer is 1.30.  the answer would be two if the concentration was 0.01M.  you just take the negative log of the concentration of hydronium ion.
108007
I think it goes like this:\n20% = .2, 25% = .25\n\nlet x = final volume\n.25(12) = .2(x)\n1.25(12) = x\n15 = x\nif x = final volume, volume added = x - 12\nor 15\n15 - 12 = 3\nadd 3 L
964564
Rate of effusion for gas 1 / rate of effusion for gas 2 = sqrt M2 / sqrt M1 (where M is molar mass)
1187917
20\n\none half of what 50 is 10 less than = 60/2 = 30\n\n20 is 10 more than 10, and 10 less than 30
426775
b = 2g - 10\ng = (b + 10)/2
71990
it would be 461/2 gallons
823484
First, make a balanced equation:\n\n2H2 + 02 → 2H2O\n\nNow find the theoretical yield (use O2 because it's the limiting reagent):\n\n10.0 g O2 * (1 mol O2/31.9988 g O2) = 0.313 mol O2\n\n0.313 mol O2 * (2 mol H2O/1 mol O2) = 0.626 mol H2O\n\n0.626 mol H2O * (18.009 g/1 mol H2O) = 11.3 g H2O (this is the theoretical yield)\n\nPercent yield = (actual yield/theoretical yield) * 100\n\nPercent yield = (7.81/11.3) * 100\n\nPercent yield = 69.1%\n\n\n* Please vote my answer as the best since I went into greater detail.  Thanks.
601468
The answer is (d) if this is the question:\n\nInitially the tank is full.  The outlet pipe is opened.  When the water in the tank is 3/4th full, both inlet pipes are opened.  Calculate the fractional level of the water in the tank after 2:05 hr.\n\nSince inlet A and outlet C cancel each other out at 50L/hr, only need to consider inlet B at 25L/hr.  Since 2:05 hr is the same as 2.083 hr (5 min = 0.083 hr), the amount of water added by inlet B is 25L/hr * 2.083 hr = 52.08 L.  Will use 52 L in the calculation.\nThe total amount of water in the tank after 2:05 hr is the initial 750 L + 52 L = 802 L.\nThe fraction level of the water is 802 L / 1000 L = 0.802\nSince 4/5 = 0.8 the answer is (d).
115336
Well first lets get the weights figured out,\n\n9% soloution,\n\nweight= (9.57 lbs/ gallon) * 6 gallons\n      = 57.42 lbs\n\n0.5% soloution,\n\nweight = (8.34 lbs/gallon) * 60 gallons\n       =500.4 lbs\n\nTotal weight = 500.4 lbs + 57.42 lbs\n             = 557.42 lbs\n\n\nnow 9% of 57.42 lbs= 5.1678 lbs\n    0.5% of 500.4 lbs= 2.502 lbs\n           Total = 2.502 +5.1678\n                 = 7.6698 lbs\n\nso to get the percentage,\n\n7.6698 lbs / 557.42 lbs\n=0.01376\n=1.376% (approximatley)\n\nSo the conecntration is approx 1.4%\n\nhope that helps
234254
sqrt of 2.5% = 1.58% (as area is 2 dimensional but radius is one-dimensional)