Answer:
first we add the same direction. 12N + 32 N=44N .
then we add the forces. 54 up + 44N down= 10N up
Scene B depicts chemical change in matter at atomic change.
Composition distinguishes a chemical reaction from a physical reaction. In a chemical process, the makeup of the components changes; in a physical change, the appearance, smell, or straightforward exhibition of a sample of matter changes without changing its composition. Despite the fact that we refer to them as physical "reactions," nothing is actually changing. A change in the substance in question's elemental composition is necessary for a reaction to occur. Therefore, from now on, we will simply refer to bodily "reactions" as physical changes.
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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.
As we know that Molarity is given as,
M = moles / V
Solving for V,
V = moles / M ------------------(1)
Also, moles is equal to,
moles = mass / M. mass -------------(2)
puting value of moles from eq. 2 into eq. 1,
V = (mass / M.mass) / M
Putting values,
V = (45 g / 164 g/mol) / 1.3 mol/dm³
V = 0.21 dm³
