1 mole H2O = 18.02g
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water is H2O ... 2 x 1.008 + 1 x 15.996 = 18.012 g/mole = 18.01 grams
First you find the Molar Mass of water using the values on the periodic table.2 Hydrogen + 1 Oxygen2*1.01 + 16.00 = 18.02 g/molThen using factor-label to cancel out units(100g H2O)*(1 mol H2O / 18.02 g H2O) = 5.54938957 molSo the number of moles in 100g of H2O is about 5.55 moles.Hope that helps.
2H2 + O2 ===> 2H2O50 grams H2 x 1 mole H2/2 g = 25 moles H2300 grams O2 x 1 mole O2/32 g = 9.375 moles O2Limiting reactant is O2, so maximum moles of H2O formed = 9.375 O2 x 2H2O/1 O2 = 18.75 molesgrams H2O = 18.75 moles H2O x 18 g/mole = 337.5 g H2O = 340 g (to 2 significant figures)
The percentage of silver nitrate by weight is 100[5.48/(5.48 + 25.0)] = 18.0 %, to the justified number of significant digits.
Assuming that the water is produced by complete combustion of propane, the balanced equation is: C3H8 + 5 O2 = 3 CO2 + 4 H2O. The gram molecular mass of propane is three times the gram atomic mass of carbon plus eight times the gram atomic mass of hydrogen = 3(12.011) + 8(1.008) = 44.097, and the gram molecular mass of water is the gram atomic mass of oxygen plus twice the gram atomic mass of hydrogen = 15.999 + 2(1.008) = 18.015. The balanced reaction equation shows that four molecules of water are produced for each molecule of propane, so that the ratio of grams of propane reacted to grams of water produced is 44.097/(4 X 18.015) = 0.61191, to the justified number of significant digits. Therefore, the amount of propane required to produce 8.00 grams of water = 8.00(0.61191) = 4.90 grams to the clearly justified number of significant digits, matching the three significant digits given for 8.00, or 4.895 grams, where the last digit is depressed because it may well not be accurate within + 1
This is a proportion problem. Call the unknown mass in grams of compound m. Then 4.50/18.3 = m/100; m =450/18.3 = 24.6 g, to the justified number of significant digits.
5 grams water is 1 teaspoon
I assume you mean 30o Celsius. Use this formula.q(joules) = mass * specific heat * change in temperatureq = (15 grams water)(4.180 J/gC)(40o C - 30o C)= 627 joules==========( perhaps 630 joules to be in significant figures territory )
The most accurate answer is 0.2 mol/kg.Related Information:The number of significant figures in the final answer must be based on the given with the least number of significant figures which is 2L.
The answer depends on whether you are measuring the drops from a slow drip or the number of drops of water in an ocean!
1. All non-zero numbers are significant 2. Zeros between non-zero digits count 3. Zeros at the beginning of a number do not count 4. Zeros at the end of a number that does not have a decimal are not significant 5. Zeros at the end of a number with a decimal are significant
Pure water can roughly hold 0.073 grams per cm at 20 degree Celsius.Impurities can change the figures.
Pure water can roughly hold 0.073 grams per cm at 20 degree Celsius.Impurities can change the figures.
2.000 = 2,000.000g 6.5 determines only one significant figure 2,000 + 6.5 + 47.546 = 2,054.046 which would be 2,054.0 In scientific notation that would equal 2.0540 * 103
76231865367 water
First you find the Molar Mass of water using the values on the periodic table.2 Hydrogen + 1 Oxygen2*1.01 + 16.00 = 18.02 g/molThen using factor-label to cancel out units(100g H2O)*(1 mol H2O / 18.02 g H2O) = 5.54938957 molSo the number of moles in 100g of H2O is about 5.55 moles.Hope that helps.
using water displacement because it doesnt float on water and it doesnt dissolve in water
The atomic mass of hydrogen is 1.008 and the molecular mass of water, with formula H2O, is 18.015. Therefore, the mass of hydrogen to that of water has the ratio of 2(1.008)/18.015 = about 0.1119, and the answer to the problem is 300/0.1119 = 2.68 X 103 grams, to the justified number of significant digits.