When zinc or aluminum was allowed to react with the copper sulfate , what was the limiting reagen?

1 answer:

5 0

Usually it is the CuSO4 that is the limiting reagent.

<span>if all of the color of the solution was gone, but there was still some zinc metal mixed in with the copper metal produced, then Zn is the excess reagent </span>

<span>f all of the color of the solution was not gone, but there was no zinc metal left in with the blue copper solution , then Zn is the limiting reagent Hope this helps.</span>

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Which solution has the lowest pH? A. 1.0 M HNO2 B. 1.0 M CH3COOH C. 1.0 M HCOOH D. 1.0 M HNO3 E. 1.0 M HPO4–

denpristay [2]

The parameter pH is a measure of the substance's acidity or basicity. It is quantitatively equal to the negative logarithm of the concentration of H+ ions. So, the lower the pH, the more acidic the substance, Otherwise, the higher the pH the more basic the substance. The pH range runs from 1 to 14, with 7 being neutral.

So, if we are asked to distinguish which of those have the lowest pH, we have to know the moles of H+ ions. Since all of them have a concentration of 0.1 M, concentration is not a factor. Thus, we just have to identify the strongest acid among the list. That is easy to answer because you only have to remember 7 strong acid occurring in nature: HCl, H₂SO₄, HNO₃, HClO₄, HClO₃, HBr, and HI. Since only HNO₃ is included in the list among the choices, the answer would be letter D.

8 0

3 years ago

Upon combustion, a compound containing only carbon and hydrogen produces 2.67 g<img src="https://tex.z-dn.net/?f=CO_%7B2%7D" id=

Hoochie [10]

Answer:

$\boxed{\text{CH$_{2}$}}$

Explanation:

1. Calculate the mass of each element

$\text{Mass of C} = \text{2.67 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.7286 g C}\\\\\text{Mass of H} = \text{1.10 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.1231 g H}$

2. Calculate the moles of each element

$\text{Moles of C = 0.7286 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.060 67 mol C}\\\\\text{Moles of H = 0.1231 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.1221 mol H}$