Answer:
No, ΔE does not always equal zero because it refers to the systems internal energy, which is affected by heat and work
Explanation:
According to the first law of thermodynamics, energy is neither created nor destroyed. This implies that the total energy of a system is always a constant.
So, according to the first law of thermodynamics we have that ΔE = q + w. This means that the value of ΔE depends on q (heat) and w(work). Hence ΔE is not always zero since it depends on the respective values of q and w.
The number that represents the coefficient on the product side of the chemical reaction, 
is 7.
<h3>Coefficients of chemical equations</h3>
In equations representing chemical reactions, the coefficient of each reactant or product of a reaction is the number that comes on the left-hand side just before the chemical formula.
The coefficient of each species in a chemical reaction is obtainable when the equation of the reaction is balanced.
For example, in the following equation: 2A + B = 3C + D
The coefficients of A, B, C, and D are 2, 1, 3, and 1 respectively.
Applying this to the product side of a chemical reaction;
It means that the coefficient of the product is 7.
More on coefficients of chemical equations can be found here: brainly.com/question/28294176
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Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.
Answer:
Adding sodium or potassium hydroxide in amounts sufficient to convert all the H2SO4 into Na2SO4 would approximately neutralize the solution. The error would be the result of the imbalance between the basicity of the hydroxide and the acidity of the bisulfate (HSO4) anion. An adjustment in concentration would have to be made to achieve an accurate approximate pH of 7. But then you didn’t ask how much we would need to add.
Explanation:
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