Answer:
Explanation:
19) it is 3d10 instead of 4d10
20) it is missing 3p6, and 4s2 before 3d5
21) Ra is not a noble gas
22) Cs is not a noble gas
Answer:
pH of HNO₃ having an hydrogen ion concentration of 0.71M is 0.149
Explanation:
HNO₃ (aqueous) ⇄ H⁺ + NO3⁻
The pH is defined as the negative log of the hydrogen ion concentration
pH = - log [H⁺]
From the question, the hydrogen ion concentration is given as 0.71M, therefore
pH = -log [0.71]
= 0.149
The unbalanced equations are the equations with different atomic numbers on the sides of the reaction. The unbalanced reaction is Na + Cl₂ → NaCl
<h3>What are balanced equations?</h3>
Balanced equations are the chemical reaction representation that has an equal number of the atomic number of the same species on the left and the right side of the reaction.
An unbalanced equation between sodium metal and chloride can be shown as:
Na + Cl₂ → NaCl
The equation is unbalanced as the number of chloride ions is more on the reactant side than the product side.
The balanced reaction will be:
2 Na + Cl₂ 2NaCl
Learn more about balanced equations here:
brainly.com/question/10306791
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Answer: The final volume of this solution is 0.204 L.
Explanation:
Given: Molarity of solution = 2.2 M
Moles of solute = 0.45 mol
Molarity is the number of moles of solute present divided by volume in liters.
Substitute the values into above formula as follows.
Thus, we can conclude that the final volume of this solution is 0.204 L.
Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
