Answer:
Explanation:
Step 1. Determine the cell potential
<u> E°/V </u>
2×[Cr ⟶ Cr³⁺ + 3e⁻] 0.744 V
<u>3×[Cu²⁺ + 2e⁻ ⟶ Cu] </u> <u>0.3419 V
</u>
2Cr + 3Cu²⁺ ⟶ 3Cu + 2Cr³⁺ 1.086 V
Step 2. Calculate ΔG°
<em>It is beneficial because it provides stronger support for the evidence related to the discovery</em>
<u>Answer:</u> <em>The correct answer is the second option that is given with shows it is beneficial.</em>
<u>Explanation:</u>
The discovery of an element whose properties are known from before would make it easier for the scientists to know the properties of elements and its uses after the discovery of element.
Since the elements’ properties are already known as element that possesses these properties and will look forward to find that and once the element is found the element can be named and can be used directly, since its uses are already known.
Answer:
There are several ways that scientists communicate our results, including written reports and scientific journal publications, and by giving presentations to our colleagues and the public. One popular venue for scientists to present to colleagues is at scientific conferences.
Explanation:
Answer:
-2.80 × 10³ kJ/mol
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.
Qcal + Qcomb = 0
Qcomb = - Qcal [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ
where,
C: heat capacity of the calorimeter
ΔT: change in the temperature
From [1],
Qcomb = - Qcal = -29.2 kJ
The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:
ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol
