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lana66690 [7]
3 years ago
12

What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii) nitrate for 30.6 seconds?

Chemistry
1 answer:
asambeis [7]3 years ago
3 0
Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Putting Values,
 
                            W  =  (10 A × 30.6 s × 31.77 g) ÷ 96500

                            W  =  0.100 g

Result:
           0.100 g of Cu
²⁺ is deposited.
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A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
Akimi4 [234]

Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

4 0
3 years ago
Temperate deciduous trees lose their leaves in Fall. Explain why trees in temperate rainforest and tropical rainforest don’t los
andrey2020 [161]

Answer:

So trees in temperate don't lose their leaves because the weather events aren't harsh enough.

Trees in tropical rainforest don't lose their leaves because they are a different type of tree known as evergreens that are green all year round.

Explanation:

Ok so first we'll define some things

Deciduous Trees=  Trees that lose all of their leaves for part of the year.

Trees shed their leaves trees to try and survive harsh weather events.

Temperate deciduous trees lose their leaves in fall to better survive the winter conditions of extreme cold and reduced daylight.

Temperate rainforests = An area that doesn't experience extremely cold or extremely hot temperatures or what we would call harsh weather events.

Broad-leaved trees in tropical rainforests are known evergreen, they are known as this as they are green all year round.

6 0
3 years ago
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7 0
3 years ago
What is the molarity of formaldehyde in a solution containing 0.25 grams of formaldehyde per mL?
Vikentia [17]

Answer:

8.33mol/L

Explanation:

First, let us calculate the molar mass of of formaldehyde (CH2O). This is illustrated below:

Molar Mass of CH2O = 12 + (2x1) + 16 = 12 + 2 + 16 = 30g/mol

Mass of CH2O from the question = 0.25g

Number of mole CH2O =?

Number of mole = Mass /Molar Mass

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Now we can calculate the molarity of formaldehyde (CH2O) as follow:

Number of mole of CH2O = 8.33x10^-3mole

Volume = 1mL

Converting 1mL to L, we have:

1000mL = 1L

Therefore 1mL = 1/1000 = 1x10^-3L

Molarity =?

Molarity = mole /Volume

Molarity = 8.33x10^-3mole/1x10^-3L

Molarity = 8.33mol/L

Therefore, the molarity of formaldehyde (CH2O) is 8.33mol/L

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Answer: I believe it’s the last one I am so sorry if I’m wrong you sound urgent and I wanted to help so I read up on wiki

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5 0
3 years ago
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