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Brums [2.3K]
1 year ago
8

How many moles of Cu are needed to react with 5.8 moles of AgNO3? Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag

Chemistry
1 answer:
Tanzania [10]1 year ago
3 0

Answer:

moles of Cu° needed =  ½ (5.8 moles) = 2.9 moles Cu° needed.

Explanation:

Cu + 2AgNO₃  =>  Cu(NO₃)₂ + 2Ag

? moles Cu + 5.8 moles AgNO₃ =>    Cu(NO₃)₂ + 2Ag

<u>Note coefficient of Cu° vs coefficient of AgNO₃ </u>

=> 1 mole Cu° < 2 moles AgNO₃ ...

=> Since moles of Cu° are smaller than moles of AgNO₃ in the given equation, the moles of Cu° needed to react with 5.8 moles AgNO₃ will be smaller than the 5.8 by the ratio of coefficients that will make 5.8 smaller. That is ...

moles of Cu° needed =  ½ (5.8 moles) = 2.9 moles Cu° needed.

Note: Using 2/1(5.8) will make value greater than 5.8 and incorrect.

One can also set up a ratio relationship as follows...

1 mole Cu° <=> 2 moles AgNO₃ (fm equation)

? mole Cu° <=> 5.8 moles AgNO₃ (problem)

=> (1 mole Cu°)/X=(2 mole AgNO₃)/(5.8 moles AgNO₃)

=> X = (1 mole Cu°)(5.8 mole AgNO₃)/2 mole AgNO₃

        = ½(5.8) mole Cu° = 2.9 mole Cu°

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Aleks04 [339]

Answer:

0.036 M

Explanation:

To do this, let's mark the dye as D and bleach as B.

We have the concentrations of both, and we already know that they react in a 1:1 mole ratio. The total volume of reaction is 9 + 1 = 10 mL or 0.010 L, and we hava both concentrations.

The problem already states that the dye reacts completely, so this is the limiting reagent, while bleach is the excess.

To know the remaining amount of bleach, we need to do this with the moles. First, let's calculate the initial moles of D and B:

moles D = 3.4x10⁻⁵ * 0.009 = 3.06x10⁻⁷ moles

moles B = 0.36 * 0.001 = 3.6x10⁻⁴ moles

Now that we have the moles, and that we know that all the dye reacts completely, let's see how many moles of bleach are left:

moles of B remaining = 3.6x10⁻⁴ - 3.06x10⁻⁷ = 3.597x10⁻⁴ moles

These are the moles presents of B after the reaction has been made. The concentration of the same will be:

[B] = 3.597x10⁻⁴ / 0.010

[B] = 0.0357

With 2 SF it would be:

[B] = 3.6x10⁻² M

5 0
1 year ago
A student determines measures the mass of one mole of carbon and finds it to be 12.22 grams. if the accepted value is 12.11 gram
Gre4nikov [31]

Answer:-

0.91% is the students % of error

Explanation: -

Accepted value= 12.11 grams

Measured value = 12.22 grams

Error = 12.22-12.11 = 0.11 grams

Percentage error = \frac{0.11 grams}{12.11 grams}x100

                           = 0.91 %

Thus 0.91% is the students % of error

5 0
1 year ago
If an electronic balance reports an object weighing 35.9920g what will be displayed when
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Answer:

2.0000 g

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It will show the number out to the 4th decimal place.  Exactly 2 g will be displayed as 2.0000 g.

8 0
2 years ago
When a 3.22 g sample of an unknown hydrate of sodium sulfate, na2so4 ⋅ h2o(s), is heated, h2o (molar mass 18 g) is driven off. T
Mariana [72]

The value of X is 10 hence the formula of unknown hydrate   sodium sulfate is  NaSO4.10 H20

calculation

step 1:find the moles of NaSO4 and the moles of H2O

moles= mass/molar mass

moles of Na2SO4=1.42÷142=0.01 moles

moles of H20=  mass of H2O/molar mass of H2O

mass of H2O= 3.22-1.42=1.8g

mole of H2O is therefore 1.8÷18=0.1 moles

step 2: find the mole ratio by dividing each mole by smallest number of mole (0.01)

that is Na2So4= 0.01/0.01 =1

            H2O= 0.1/0.01=10

6 0
1 year ago
iron combines with 4.00 g of copper (11) nitrate to form 6.01 g of Iron (l) nitrate and 0.400 g copper metal. how much iron did
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Answer:

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Explanation:

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7 0
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