18 grams of water is 1 mole water.
Water is H2O.
That means there's 2 moles H and 1 Mole O.
We're looking for "O", and there's one 1.
So find the molar mass of O. (which is approx. 16)
so 16 * 1 = 16.
There are 7.24 x 10^23 atoms of oxygen in 13 grams of water. This calculation is based on the molar mass of water (18.015 g/mol) and the fact that each water molecule contains two hydrogen atoms and one oxygen atom.
The balanced chemical equation for the reaction between oxygen and hydrogen is2H2 + 02 -> 2H2OThus 2.2 moles of oxygen reacts with 4.4 moles of hydrogen to form 4.4 moles of steam (water in gaseous state).The mass of H2O obtained is thus 4.4 x 18.0 = 79.2g.
100 g water dissolve 45,8 g potassium chloride at 50 o 0C.
There are 0.47 grams of oxygen in 0.53 grams of NaHCO3.
1 mole of water is equal to 18.02 grams. H = 2 * 1.01 g = 2.02 g O = 1 * 16.00 g = 16.00 g Total = 18.02 g 1.8 g H2O * (1 mol H2O/18.02 g H2O) = 0.099889 mol H2O There are about 9.99*10-2 moles in 1.8 grams of water.
7
HCN is hydrocyanic acid and contains no oxygen, so producing it doesn't require oxygen. So zero grams.
For the reaction 2H₂ + O₂ → 2H₂O, we know that the molar ratio of H₂ to O₂ is 2:1. To produce 900 grams of water, we need 450 grams of hydrogen (900g / 2). Therefore, we need to add 450 grams of hydrogen to 800 grams of oxygen to produce 900 grams of water.
2.2 mol water = 2.2 (mol) * 18 (g/mol) water = 39.6 (mol*g/mol) = 40 g18 g/mol = mol mass of H2O = 2*H + 1*O = (2*1 + 16) g/mol
One McDonald's Filet-O-Fish has 400 calories and 18 grams of fat in it.
In the reaction of Fe with H₂O to produce FeO and H₂, the ratio is 3 moles of H₂O for every 1 mole of Fe. Since the molar mass of Fe is 56 g/mol and that of H₂O is 18 g/mol, we use these values to find that 750 grams of Fe will produce 750 * (3 * 18 / 56) = 857.14 grams of H₂O.
To determine the grams of glucose formed, we can use the balanced equation for photosynthesis: 6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂. From the equation, 6 grams of carbon dioxide and 6 grams of water produce 1 gram of glucose and 6 grams of oxygen. Given that 16 grams of oxygen are produced, that corresponds to 16 grams of O₂, which comes from 6 grams of CO₂ and 6 grams of H₂O multiple times. Thus, when 22 grams of CO₂ and 9 grams of H₂O are used, we can find that glucose produced is 6.67 grams.
As the molar mass of water is 18 g/mol (1*2 H + 16 O) 10 moles*18 g/mole=180 grams.
To teach you how to do this problem: 1. Look at the Periodic Table. O =16 grams/mole, H =1.01 grams/mole 2. Calculate the Molecular Weight of Water. 1*O + 2*H: 16+2.02=18.02g/mole 3. Divide the mass of what you have by the Molecular Weight: 35.1g H2O/(18.02 g H2O/mole H2O) 4. The units g H2O cancel leaving you with mole H2O. 5. 35.1 grams H2O= 1.95 moles H2O
1 oz = 28.34 grams 1 gram = 0.03 oz
To determine the quantity of oxygen that contains nearly the same number of molecules as 36 grams of water, we first calculate the number of molecules in 36 grams of water (H₂O). There are approximately 20 moles of water in 36 grams (since the molar mass of water is about 18 g/mol), which corresponds to about (1.2 \times 10^{24}) molecules. To find the equivalent amount of oxygen (O₂), we calculate that 20 moles of O₂ (with a molar mass of about 32 g/mol) would also contain (1.2 \times 10^{24}) molecules. Therefore, approximately 640 grams of oxygen (O₂) would contain a similar number of molecules as 36 grams of water.
An Olympic swimming pool holds about 2,500 cubic meters of water, which is equivalent to 2,500,000 liters. Since one mole of water (H₂O) is approximately 18 grams and one liter of water has a mass of about 1,000 grams, this means there are roughly 2,500,000 moles of water in an Olympic swimming pool. Thus, you can estimate that there are around 138,888,889 moles of water in such a pool.