C + O2 ==> CO2At STP 1 mole CO2 = 22.4 L
10L x 1 mole/22.4 L = 0.446 moles CO2 needed
1 mole C = 1 mole CO2
Therefore you need 0.446 moles C
grams C = 0.446 moles C x 12 g/mole = 5.36 grams = 5 g (to 1 significant figure)
To prepare a 3% solution of sulfosalicylic acid, you would need 30 grams of sulfosalicylic acid for every 1 liter of solution.
If 12 grams of carbon were used to form the 22 grams of carbon dioxide, this implies that 12 grams of oxygen were consumed in the reaction. Since 20 grams of oxygen were initially available, only 8 grams of oxygen are left unused.
There are approximately 3.633 x 10^23 carbon atoms in 21.84 grams of carbon.
A 2-liter bottle of Sprite contains about 5.3 grams of carbon dioxide.
A 12-ounce can of Sprite contains approximately 19 grams of carbon dioxide, which is used to carbonate the beverage.
To prepare a 3% solution of sulfosalicylic acid, you would need 30 grams of sulfosalicylic acid for every 1 liter of solution.
85.636 grams carbon (1 mole C/12.01 grams) = 7.1304 moles of carbon ---------------------------------
If 12 grams of carbon were used to form the 22 grams of carbon dioxide, this implies that 12 grams of oxygen were consumed in the reaction. Since 20 grams of oxygen were initially available, only 8 grams of oxygen are left unused.
To prepare 0.5L of D5W (5% dextrose in water), you need 25 grams of dextrose. This is because 5% of 0.5L is 25 grams.
To calculate the grams of KI required, use the formula: grams = molarity (M) × volume (L) × molar mass (g/mol). The molarity is 0.04 M and the volume is 0.5 L (500 ml). The molar mass of KI is approximately 166 g/mol. Therefore, grams of KI required = 0.04 M × 0.5 L × 166 g/mol = 3.32 grams.
31.8 grams carbon (1 mole C/12.01 grams)(6.022 X 1023/1 mole C)= 1.59 X 1024 atoms of carbon===================
11 grams because all is reacted and there is no reactant left over, although if there were only 3 grams of carbon there would have to be 6 grams of oxygen for this to be viable as carbon dioxide is CO2 so the question asked was itself wrong.
There are approximately 3.633 x 10^23 carbon atoms in 21.84 grams of carbon.
To prepare a 5% NaCl solution, you will need 200 grams of NaCl for 4000 mL (4 L) of solution. This is calculated as 5% of 4000 mL, which equals 200 grams.
Balanced equation. C + 2Cl2 -> CCl4 10 grams carbon (1 mole C/12.01 grams)(1 mole CCl4/1 mole C)(153.81 grams/1 mole CCl4) 128 grams carbon tetrachloride produced -----------------------------------------------------
4314.9 grams
17 grams carbon (1 mole C/12.01 grams)(6.022 X 1023/1 mole C) = 8.5 X 1023 atoms of carbon =====================