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avanturin [10]
4 years ago
9

How many gallons of a 60​% antifreeze solution must be mixed with 70 gallons of 25​% antifreeze to get a mixture that is 50​% ​a

ntifreeze? Use the​ six-step method.
Chemistry
1 answer:
Neporo4naja [7]4 years ago
6 0

Answer:

The answer to your question is 175 gallons

Explanation:

Data

60% antifreeze solution  volume = ? = x

25% antifreeze solution   volume = 25%

Final concentration 50%

Process

              0.6x + 0.25(70) = 0.5(70 + x)

Simplification

              0.6x + 17.5 = 35 + 0.5x

              0.6x - 0.5x = 35 - 17.5

                          0.1x = 17.5

                              x = 17.5/0.1

Result

                              x = 175 gallons

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1. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hour
lianna [129]

*Answer:

Option A: 59.6

Explanation:

Step 1: Data given

Mass of aluminium = 4.00 kg

The applied emf = 5.00 V

watts = volts * amperes

Step 2: Calculate amperes

equivalent mass of aluminum = 27 / 3 = 9  

mass of deposit = (equivalent mass x amperes x seconds) / 96500

4000 grams = (9* amperes * seconds) / 96500

amperes * seconds = 42888888.9

1 hour = 3600 seconds

amperes * hours = 42888888.9 / 3600 = 11913.6

amperes = 11913.6 / hours

Step 3: Calculate kilowatts

watts = 5 * 11913.6 / hours

watts = 59568 (per hour)

kilowatts = 59.6 (per hour)

The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V

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Write balanced chemical equation for the reactions used to prepare each of the following compounds from the given starting mater
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                N_{2} + 3H_{2} \rightarrow 2NH_{3}

So here, oxidation state of nitrogen is changing from 0 to -3. Whereas oxidation state of hydrogen is changing from 0 to +3.

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Since, it is a combination reaction. Therefore, no change in oxidation state of reacting species is taking place.

(b)  Reaction for conversion of liquid bromine to hydrogen bromide is as follows.

            H_{2}(g) + Br_{2}(l) \rightarrow 2HBr(g)

Here, oxidation state of hydrogen is changing from 0 to +1. Therefore, oxidation of hydrogen is taking place. On the other hand, oxidation state of bromine is changing from zero to -1. Therefore, reduction of bromine is taking place.

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               Zn(s) + S(s) \rightarrow ZnS(s)    

Here, oxidation state of zinc is changing from 0 to +2. Hence, oxidation of zinc is taking place.

Oxidation state of sulfur is changing from 0 to -2. Therefore, reduction of sulfur is taking place.

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