Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
Answer:
2.97 × 10¹³ g
Explanation:
First, we have to calculate the biomass the is burned. We can establish the following relations:
- 2.47 acre = 10,000 m²
- 10 kg of C occupy an area of 1 m²
- 50% of the biomass is burned
The biomass burned in the site of 400,000 acre is:
Let's consider the combustion of carbon.
C(s) + O₂(g) ⇒ CO₂(g)
We can establish the following relations:
- The molar mass of C is 12.01 g/mol
- 1 mole of C produces 1 mole of CO₂
- The molar mass of CO₂ is 44.01 g/mol
The mass of produced is CO₂:
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.
Answer: 2 atoms
Explanation: 2 in Co2 means 2 atoms
