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aliina [53]
4 years ago
8

You are in Paris and want to buy peaches for lunch. The sign on the fruit stand indicates that peaches cost 11.50 euros per kilo

gram. On that day, the euro is 1.25 dollars. How much would a pound of peaches cost in dollars?
Chemistry
1 answer:
nydimaria [60]4 years ago
4 0

Answer:

1 lb of peaches costs 6.52 dollars

Explanation:

You can solve this problem using the rule of three.

You might want to follow these steps:

1. Analyze the giving data

2. Organize it and write the proportions  

3. Clear the x value

4. Solve by multiplying or dividing

Now for the exercise:

Data:

$11.50 euros per 1 kg

1 euro is 1.25 dollars

Conversion factor 1 kg = 2.02462 lb

The problem is asking how much will be the cost in dollars to buy 1 lb of peaches? As you can see! You need to change the units to modify the data information.

Remember that 1 kg = 2.02462 lb

First change the kg to lb, in this case use directly the conversion factor:

2.02462 lb =  1 kg

Now you want to know the cost of 1 lb

1 lb               ---->   x euros?

2.20462 lb -----> 11.50 euros

Clear the incognite x euros?

1 lb (11.50 euros) =   x euros? (2.20462 lb)

x euros? = 1 lb (11.50 euros) / 2.20462 lb = 5.22 euros

The cost of 1 lb of peaches will be 5.22 euros, but you need to know the cost in dollars.

You will need to write another proportion:

1 euro          -----> 1.25 dollars

5.216 euros -----> x dollars?

Now solve for incognite x dollars?

1 euro (x dollars?) = 1.25 dollars (5.216 euros)

x dollars? = 1.25 dollars (5.216 euros)/(1 euro) = 6.52 dollars

Now you that 1 lb of peaches costs 6.52 dollars

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Answer:

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

Explanation:

Cu + HNO3 → Cu(NO3)2 + NO + H2O

The first step is to write the oxidation numbers for each atoms in the given equation  

Cu0 + H+1N+5O-23 → Cu+2(N+5O-23)2 + N+2O-2 + H+12O

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H+1N+5O-23 + 3e- + 3H+ → N+2O

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Make electron gain equivalent to electron lost.

3Cu0 + 6HNO3 → 3Cu+2(N+5O-23)2 + 6e- + 6H+

2H+1N+5O-23 + 6e- + 6H+ → 2N+2O-2 + 4H2O

Complete reaction  

3Cu0 + 8H+1N+5O-23 + 6e- + 6H+ → 3Cu+2(N+5O-23)2 + 2N+2O-2 + 6e- + 4H2O + 6H+

Simplify the equation

3Cu0 + 8H+1N+5O-23 → 3Cu+2(N+5O-23)2 + 2N+2O-2 + 4H2O

Final equation  

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

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