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natita [175]
2 years ago
11

Introduction:

Chemistry
2 answers:
Alex777 [14]2 years ago
8 0

Answer:

............................................

ki77a [65]2 years ago
8 0

Answer: im not sure if this is the best answer but i wrote, "This exploration is about how the volume of a gas varies with temperature, and what the effect does to the gas and what effect it has on the results."

Explanation: hope this helps.

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The temperature of a freezer started at 18şc. After cooling for a few hours, the freezer had a temperature of –12şc. What is the
andre [41]
20 minus 2 plus 1 equals negative -2000000
8 0
2 years ago
A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

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.

4 0
3 years ago
Arizona was the site of a 400,000-acre wildfire in June 2002. How much carbon dioxide (CO2) was produced into the atmosphere by
kodGreya [7K]

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:

400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC

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:

8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}

4 0
3 years ago
The solubility of NaCH3CO2 in water is ~1.23 g/mL. What would be the best method for preparing a supersaturated NaCH3CO2 solutio
Len [333]

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.

5 0
3 years ago
How many atoms of oxygen is in CO2?
rosijanka [135]

Answer: 2 atoms

Explanation: 2 in Co2 means 2 atoms

7 0
3 years ago
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