Answer:
The concentration of the copper (II) sulfate solution is 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM
Explanation:
The concentration of a solution is the amount of solute dissolved in a given volume of solution. In this case, the concentration of the copper(II) sulfate solution in micromoles per liter (symbol ) is the number of micromoles of copper(II) sulfate dissolved in each liter of solution. To calculate the micromoles of copper(II) sulfate dissolved in each liter of solution you must divide the total micromoles of solute by the number of liters of solution.
Here's that idea written as a formula: c= n/V
where c stands for concentration, n stands for the total micromoles of copper (II) sulfate and V stands for the total volume of the solution.
You're not given the volume of the solution in liters, but rather in milliliters. You can convert milliliters to liters with a unit ratio: V= 150. mL * 10^-3 L/ 1 mL = 0.150 L
Next, plug in μmol and liters into the formula to divide the total micromoles of solute by the number of liters of solution: c= 31 μmol/0.150 L = 206.66 μmol/L
Convert this number into scientific notation: 2.06 * 10^2 μmol/L or 2.06 * 10^2 μM
Answer:
Project 3.
Explanation:
Project 3's anticipated cost is 12 to 17 million dollars. It is a <em>lower </em>anticipated cost than Project 2 and Project 4, but <em>higher</em> than Project 1 by one million dollars at maximum cost anticipation. Additionally, the percentage of wildlife to benefit is 70-80%, which is <em>second</em> to the most wildlife to benefit which is Project 4 at 75-80%.
And finally, for community support for Project 3 - the chart lists it as high. This outclasses Project 2 and Project 4, but balances with Project 1. However, Project 1 costs 13 to 16 million dollars and <em>only</em> benefits 15-25% of wildlife.
Explanation:
"Compound: A substance that is made up of more than one type of atom bonded together.
Mixture: A combination of two or more elements or compounds which have not reacted to bond together; each part in the mixture retains its own properties."
- Libre Texts
Answer : The correct option is, (B) 
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
..........(1)
where,
= rate of effusion of unknown gas = 
= rate of effusion of oxygen gas = 
= molar mass of unknown gas = ?
= molar mass of oxygen gas = 32 g/mole
Now put all the given values in the above formula 1, we get:
The unknown gas could be carbon dioxide 
that has approximately 44 g/mole of molar mass.
Thus, the unknown gas could be carbon dioxide
C. dim light is <span>the answer
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