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:
tritium and deuterium are combined and result in the formation of helium
You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.
Explanation:
To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:
2 Mg (s) + O₂ (g) → 2 MgO + heat
CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat
However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).
Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)
2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)
So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.
we have used the following notations:
(s) - solid
(g) - gas
(l) - liquid
Learn more about:
combustion reactions
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Answer:
91383 J
Explanation:
The equation of the reaction can be represented as:
------>
Given that:
The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.
The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.
= 
where:
= enthalpy of reaction
= the difference in the heat capacities of the products and the reactants.
∴
=

= ![1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'](https://tex.z-dn.net/?f=1%2891300%20J.mol%5E%7B-1%7D%20%29%20%2B%5Cint%5Climits%5E%7B435%7D_%7B298.15%7D%20%5B%7B%2829.86%29-%5Cfrac%7B1%7D%7B2%7D%2829.38%29-%5Cfrac%7B1%7D%7B2%7D29.13%7D%5DJ.K%5E%7B-1%7D.mol%5E%7B-1%7D%20%5C%2C%20dT%27)
= 91300 J + (0.605 J.K⁻¹)(435-298.15)K
= 91382.79 J
≅ 91383 J
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