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umka2103 [35]
3 years ago
5

Part A.4 and Part B. In a hurry to complete the experiment, Anna withdrew two volumes of solution from Part A.2 before the preci

pitate had settled. As a result, what dilemma might she have encountered in Part B
Chemistry
1 answer:
Harman [31]3 years ago
6 0

Answer:

Anna may have trouble because of inaccurate results or observations.

<u>Explanation:</u>

In chemistry, a precipitate is an <em>insoluble solid</em> that is gotten from a liquid solution. Because Anna withdrew two volumes of solution from Part A.2 before the precipitate had settled she may find it difficult to determine the limiting reactant; that is the chemical element or substance that limits the amount of product made during the chemical reaction because some precipitate was captured with the liquid lying above, which when adding the reagent, will not show the difference between them.

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Calculate the change in temperature that occurs when 3.50 x 10^-2kg of copper heated by 635 J of energy. Ccu = 0.385 j/gxC
vovikov84 [41]

Answer:

The change in T° is 47.1 °C

Explanation:

Calorimetry formula to solve this:

Q = m . C . ΔT

We replace the data gien:

635 J = 3.50×10⁻² kg . 0.385 J/g°C . ΔT

In units of C, we have g and the mass (m) is in kg. So let's convert it from kg to g → 3.50×10⁻² kg . 1000 g / 1kg = 35 g. Now we can determine the ΔT:

635 J = 35 g . 0.385 J/g°C . ΔT

635 J / 35 g . 0.385 J/g°C = ΔT

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6 0
3 years ago
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Alexxx [7]

Answer:

I believe it's either a or b but I'm pretty sure that it's a

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Consider the balanced equation. 2HCl + Mg MgCl2 + H2 If 40.0 g of HCl react with an excess of magnesium metal, what is the theor
Serga [27]

Answer:

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Explanation:

Balanced equation, Mg+2HCl\rightarrow MgCl_{2}+H_{2}

As Mg remain present in excess therefore HCl is the limiting reagent.

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Molar mass of HCl = 36.46 g/mol

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So, theoretical yield of H_{2} = (0.550\times 2.016)g=1.11g

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