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soldier1979 [14.2K]
3 years ago
8

What is the percentages of Carbon, Hydrogen and oxygen in C6H12O6

Chemistry
1 answer:
lara31 [8.8K]3 years ago
6 0
First calculate the Ar of the compound.. ( Ar of C6H12O6 is 180)
Then divide the At of the required element by 180 and multiply by 100...

 so, for carbon 12 * 6 = 72 (THERE ARE 6 CARBON ATOM)
  
                            (72/180) * 100 =40%
    ITS THE SAME FOR THE OTHER 2 ELEMENT
            
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If you move a substance from one container to another and its volume changes substances is gas. 
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Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

5 0
3 years ago
Predict the electronegativity of the yet undiscovered element with Z = 119.
Lera25 [3.4K]
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We also know that the trend in the electronegativity is to decrease when you go up-down ia group.

The known electronegativities of the elements of this group are:

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Then the hypotetical element Z = 119 would probably have an electronegativity slightly below 0.70, for sure in the range 0.60 - 0.70.



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Answer:

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Explanation:

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