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klasskru [66]
3 years ago
13

How many moles of carbon in 6.64 moles of CCl2 F

Chemistry
1 answer:
Bad White [126]3 years ago
3 0

Answer: 6.64 moles of carbon.

Explanation:

Given data:

Number  of moles of C = ?

Number of moles of CCl₂F₂ = 6.64 mol

Solution:

In one mole of CCl₂F₂ there is one mole of carbon two moles of chlorine and two moles of fluorine are present.

In 6.6 moles of CCl₂F₂ :

Moles of carbon = 6.64 × 1 = 6.64 moles of carbon.

Moles of chlorine = 6.64× 2 = 13.28 moles of chlorine

Moles of fluorine = 6.64× 2 = 13.28 moles of fluorine

Read more on Brainly.com - brainly.com/question/15602143#readmore

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In an alkane the mass ratio between hydrogen and carbon is 7/36. What is the formula of alkane?
ch4aika [34]

Answer:

CnH2n+2 formula

Explanation:

Alkanes have the general formula CnH2n+2, so if an alkane had 7 carbon atoms, it would have the molecular formula C7H16.

3 0
1 year ago
A student pushes on a crate with a force of 100N directed to the right. What force does the crate exert on the student?
Kamila [148]
Hello!

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6 0
3 years ago
Which physical property of the water changes when it reaches its melting point
KiRa [710]
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6 0
3 years ago
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In Denver, Colorado the elevation is about 5,280 feet above sea level. Explain what potential effects this may have on the solub
Colt1911 [192]

Answer:

The solubility of the gaseous solute decreases

Explanation:

As we know, pressure decreases with altitude. This means that, at higher altitudes, the pressure is much lower than it is at sea level.

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Hence, in Denver, Colorado where the elevation is about 5,280 feet above sea level, a gaseous solute is less soluble than it is at sea level due to the lower pressure at such high altitude.

4 0
3 years ago
How many grams of glucose are needed to prepare 400. g of a 2.00% (m/m) glucose solution g?
Dominik [7]
The grams   of glucose  are  needed  to  prepare  400g  of  a 2.00%(m/m)  glucose  solution  g  is  calculated  as  below

=% m/m =mass  of the solute/mass  of  the  solution  x100

let mass of   solute  be represented  by  y
mass  of solution = 400 g
 % (m/m) = 2% = 2/100

 grams  of  glucose  is  therefore =2/100 =  y/400
by cross  multiplication

100y = 800
divide   both side  by  100

y= 8.0 grams



5 0
3 years ago
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