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kow [346]
3 years ago
12

Solid aluminum is placed in a beaker with a blue solution of copper(II) nitrate. The resulting reaction causes the blue solution

to turn colorless. The solid aluminum is consumed while a spongy, dark brown solid forms. Based on the observations listed above, which of the following inference(s) would be valid?
A. A synthesis reaction occurred.
B. A single replacement reaction occurred.
C. A double replacement reaction occurred.
D. A redox reaction occurred.
Chemistry
1 answer:
konstantin123 [22]3 years ago
7 0

Answer:

Option D. A redox reaction occurred.

Explanation:

From the question given above it evident that a redox reaction has occurred because there is a colour change when aluminum was introduced into the beaker containing blue solution of copper(II) nitrate.

The equation for the reaction is given below below:

2Al + 3Cu(NO3)2 —> 2Al(NO3)3 + 3Cu

The oxidation number of Al changes from 0 to +2 indicating oxidation and the oxidation number of Cu changes from +2 to 0 indicating reduction.

As the oxidation - reduction reaction occurs, a colour change is observed.

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If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did
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<u>Answer:</u>

<em>4.5 L water we have in litres (L).</em>

<em><u></u></em>

<u>Explanation:</u>

Q=m\times c \times \Delta T

where

\Delta T = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0  cal/(g℃))  

m is the mass of water

Plugging in the values  

\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$

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