I mole - 16g of methane is 1 mole. At STP it would occupy 22.4 liters
8 grams. Or more exactly, in 30.0 grams of methane there is 7.54 grams of hydrogen.
160 grams = 5.64 ounces.
To determine the grams of methane needed, we can use the principle of conservation of mass. The total mass of products (water and carbon dioxide) plus the mass of oxygen should equal the mass of the reactants (methane and oxygen). Here, we have 9 grams of water and 11 grams of carbon dioxide, totaling 20 grams of products. Since 16 grams of oxygen were used, the mass of methane needed is 20 grams (products) - 16 grams (oxygen) = 4 grams of methane.
170 grams = 0.374785846 pounds (rounded: 0.375 pounds).
160 grams = 5.64383391 ounces
0.160 kg is equal to 160 grams.
8 grams. Or more exactly, in 30.0 grams of methane there is 7.54 grams of hydrogen.
160 grams = 5.6 ounces.
160 grams = 5.64 ounces.
This is very basic math. Carbon = 6/8*Methane (6/8)*4=3 3 g
.160 ounces = 4.535 grams.
160 grams = about 5.644 ounces.
170 grams = 0.374785846 pounds (rounded: 0.375 pounds).
160 grams = 5.64383391 ounces
160 mg = 0.16 g
To convert 160 dag (decagrams) to grams, you multiply by 10, since 1 dag is equal to 10 grams. Therefore, 160 dag is equal to 1,600 grams. In terms of kilograms, 1,600 grams is equal to 1.6 kg. Thus, 160 dag is equivalent to 1.6 kg.
To calculate the grams of dichloromethane produced, we first need to convert the mass of methane from kilograms to grams, which is 1,540 grams. Given a yield of 48.2%, we can multiply this by the yield percentage to find the actual amount of dichloromethane produced: 1,540 grams of methane x 0.482 = 742.28 grams of dichloromethane.