700 L of water was produced if 350.0 L of carbon dioxide were made at STP.
The quantitative relationship (ratio) between reactants and products in a chemical reaction that produces gases is known as gas stoichiometry. When the created gases are presumed to be ideal and their temperature, pressure, and volume are all known, gas stoichiometry is applicable.
The ideal gas equation is PV=nRT, where n is the number of moles and R is the gas constant, P is the pressure measured in atmospheres (atm), V is the volume measured in liters (L), and
Calculations based on stoichiometry assist scientists and engineers who work in the business world in estimating the number of items they will make using a particular process. They can also assist in determining if a product will be economical to produce.
Reduced growth, reproduction, and survivability for the consumer are typically the results of a significant stoichiometric imbalance between the primary producer and consumer.
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The correct answer is option A. Energy cannot be created during an ordinary chemical reaction. There is no such thing as an ordinary chemical reaction. Energy cannot be created or destroyed this is according to the law of conservation of energy. It can only be transformed from one form to another form.
Answer:
= 331.81 g
Explanation:
Molarity is calculated by the formula;
Molarity = Moles/volume in liters
Therefore;
Moles = Molarity ×Volume in liters
= 0.35 M × 1.575 L
= 0.55125 Moles
But; Molar mass of Ba3(PO4)2 is 601.93 g/mol
Thus;
Mass = 0.55125 moles × 601.93 g/mol
<u>= 331.81 g</u>
Answer:
CH2O
Explanation:
Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.
C = 5.692/14.229 * 100 = 40%
O = 7.582/14.229 * 100 = 53.29%
H = 0.955/14.229 * 100 = 6.71%
We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.
C = 40/12 = 3.333
O = 53.29/16 = 3.33
H = 6.71/2 = 6.71
Dividing by the smaller value which is 3.33
C = 3.33/3.33 = 1
O = 3.33/3.33= 1
H = 6.71/3.33 = 2
The empirical formula of the compound ribose is CH2O
