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PolarNik [594]
2 years ago
9

Hot air is more / less dense than cooler air. Choose the correct one.

Chemistry
1 answer:
Paha777 [63]2 years ago
4 0

it is less dense then cold

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Balance each of these equations.
Oksanka [162]

Answer:

A. 2NO + O2 -> 2NO2

B. 4Co + 3O2 -> 2Co2O3

C. 2Al + 3Cl2 -> Al2Cl6

D. 2C2H6 + 7O2 -> 4CO2 + 6H2O

E. TiCl4 + 4Na -> Ti + 4NaCl

8 0
2 years ago
4P+ 5O2 -> P4O10
Ratling [72]
If 4 moles of P is used by 5 mole of O2
then....0.489 moles will be used by 5/4 × .489 = .611 moles of O2

so .611 moles

so if 4 moles of P is burnt , 1 mole of P4O10 is produced ....so for .489 moles...... .489/4=.122 moles !
so mass will be .122× 283.89 = 34.7 grams

so first ans is .611 moles and second is 34.7 grams !

if you have any problem regarding this , just comment !!!
4 0
3 years ago
Read 2 more answers
A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent
Degger [83]

Answer :  The % of (+) limonene isomer = 79%


                The % of (-) limonene isomer = 0%


                The % of enantiomeric excess = 58%


Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.


Given,


% of pure limonene enantiomer = The % of (+) limonene isomer = 79%


Therefore, The % of (-) limonene isomer = 0%


Formula used :  

\%(+)\text{ isomer}=\frac{ee}{2}+50\%


Where,         ee → enantiomeric excess


Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


     {ee}=\frac{\%(+)-50\%}{2}


            =\frac{79\%-50\%}{2}


              = 58%



7 0
3 years ago
Read 2 more answers
Đốt cháy hết 9g kim loại magie Mg trong không khí tích tụ được 15g hợp chất magie oxit MgO. Biết rằng magie cháy là xảy ra phản
vfiekz [6]

Answer:

jijji[ojooooooooooooooooooooooooo

Explanation:

kjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

5 0
2 years ago
What is the difference between S-32 and S-36?
Blizzard [7]

Answer:

its easy ask to chrome or search in yt ;)

8 0
2 years ago
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