The answer is A :) hope this help
Answer:
7.28 mol Na2SO4
Explanation:
Since it is already in moles, all we have to do is use a molar ratio
A molar ratio is the proportions of reactants and products using the balanced equation. When writing a mole ratio, the given information must cross out with the right thing.
7.28 mol H2SO4 * 1 mol Na2SO4/1 H2SO4 = 7.28 mol Na2SO4
*notice how the H2SO4 crosses out
Answer:
The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.
Explanation:
How does the concept of conservation of mass apply to chemical reactions? the reactants and products have exactly the same atoms. the reactants and products have exactly the same molecules. the change in the amount of matter is equal to the change in energy.
can someone help me with my qustions?
Actually Welcome to the Concept of the Lens and Ray Optics.
We must understand the Concept that,
1.) When a beam of parallel light pass through a Convex lens, all the rays get deviated and converges to a special point at the axis that is a Focus.
2.) When a beam of light passes through the Concave lens,the light beams diverge from one another forming a virtual focus in the primary side of lens,
Hence the above given diagram beam rays are diverging, it is a CONCAVE LENS
answer is C.) Concave lens
Answer:
b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.
Explanation:
The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).
<em>What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?</em>
<em>a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves.</em> NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.
<em>b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. </em>YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.
<em>c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.</em> NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.
