First, we need to determine the half reaction of magnesium. It would be expressed as:
Mg2+ + 2e- = Mg
Given the mass of magnesium metal that is produced, we calculate the total charge of the electrolysis by the relations we can get from the half reaction. We do as follows:
4.50 kg Mg ( 1000 g / 1 kg ) ( 1 mol / 24.305 g ) ( 2 mol e- / 1 mol Mg ) ( 96500 C / 1 mol e- ) = 35733388.2 C
We are given the applied EMF in units of V. This value is equal to J/C. So, 5 V is equal to 5 J/C.
35733388.2 C (5 J/C) = 178666941 J
178666941 J ( 1 kW-h / 3.6x10^6 J ) = 49.63 kW-h
Answer:
The answer is FALSE
Explanation:
I took the test and it was false,
also if it was true then in industrialized countries would be more eco-friendly and there wouldn't be a huge hole in our ozone layer. and no more wars over oil.
Answer:
102g
Explanation:
To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.
In this case, the given equation is already balanced.

From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.
Mole= Mass ÷Mr
Mass= Mole ×Mr
<u>Method 1: using the </u><u>mass of glucose</u>
Mr of glucose
= 6(12) +12(1) +6(16)
= 180
Moles of glucose reacted
= 200 ÷180
=
mol
Amount of ethanol formed: moles of glucose reacted= 2: 1
Amount of ethanol
= 
=
mol
Mass of ethanol
= ![\frac{20}{9} \times[2(12)+6+16]](https://tex.z-dn.net/?f=%5Cfrac%7B20%7D%7B9%7D%20%5Ctimes%5B2%2812%29%2B6%2B16%5D)
= 
= 102 g (3 s.f.)
<u>Method 2: using </u><u>mass of carbon dioxide</u><u> produced</u>
Mole of carbon dioxide produced
= 97.7 ÷[12 +2(16)]
= 97.7 ÷44
=
mol
Moles of ethanol: moles of carbon dioxide= 1: 1
Moles of ethanol formed=
mol
Mass of ethanol formed
= ![\frac{977}{440} \times[2(12)+6+16]](https://tex.z-dn.net/?f=%5Cfrac%7B977%7D%7B440%7D%20%5Ctimes%5B2%2812%29%2B6%2B16%5D)
= 102 g (3 s.f.)
Thus, 102 g of ethanol are formed.
Additional:
For a similar question on mass and mole ratio, do check out the following!
Answer:
67.1%
Explanation:
Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:
<em>Moles Na₂CO₃ - 105.99g/mol-:</em>
6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.
As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:
0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃
And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):
0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.
And percent of NaHCO₃ in the sample is:
10.06g NaHCO₃ / 15g Sample * 100 =
<h3>67.1%</h3>
Answer:
The given statement is false.
Explanation:
- A common method of experimentation that is used to collect data on hypotheses of end up causing-effect is a contrast between two groups.
- One community, the experimental group, is receiving medication for having any result. Some other group becomes left exposed, the control group, whether creating a different treatment.