I believe that this is the answer 520.03 g
This problem is providing a chemical equation between two hypothetical elements, X and Y and asks for the moles of X that are needed to produce 21.00 moles of D in excess Y. After the following work, the answer turns out to be 15.75 mol X:
<h3>Mole ratios:</h3><h3 />
In chemistry, one the most crucial branches is stoichiometry, which allows us to perform calculations with grams, moles and particles (atoms, molecules and ions). It is based on the concept of mole ratios, whereby the moles of a specific substance can be converted to moles of another one, say product to reactant, reactant to reactant, reactant to product and product to product.
<h3>Calculations:</h3>
In such a way, since 21.00 moles of D are given, we need the mole ratio of D to X in order to get the answer, which according to the reaction is 3:4 based on their coefficients in the reaction. Hence, we calculate the required as follows:
Learn more about mole ratios: brainly.com/question/15288923
The products are on the right side of the chemical equation, while the reactants are on the left side of the chemical equation.
Therefore if we have:
methane + oxygen → carbon dioxide + water
Then the reactants are methane + oxygen
The products are: A) Carbon dioxide + Water
Answer:
<u>Osmosis </u>is the process that describes water moving through a membrane.
I am not sure about the second question.
Explanation:
Answer:
2.29 m.
Explanation:
The following data were obtained from the question:
Mass of KOH = 42.3 g
Molar mass of KOH = 56.11 g/mol
Mass of water = 329 g
Molality of KOH = ?
Next, we shall determine the number of mole in 42.3 g of KOH. This can be obtained as follow:
Mass of KOH = 42.3 g
Molar mass of KOH = 56.11 g/mol
Mole of KOH =?
Mole = mass /Molar mass
Mole of KOH = 42.3/56.11
Mole of KOH = 0.754 mole
Next, we shall convert 329 g of water to kilogram (kg). This can be obtained as follow:
1000 g = 1 kg
Therefore,
329 g = 329 g /1000 g × 1 kg
329 g = 0.329 kg
Therefore, 329 g of water is equivalent to 0.329 kg
Finally, we shall determine the molality of the KOH solution ad follow:
Molality is defined as the mole of solute per unit kilogram of solvent (water) i.e
Molality = mole/ mass (kg) of water
Mole of KOH = 0.754 mole.
Mass of water = 0.329 kg.
Molality = mole/ mass (kg) of water
Molality = 0.754/0.329
Molality = 2.29 m
Therefore, the molality of the KOH solution is 2.29 m.
