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GrogVix [38]
3 years ago
6

WILL AWARD 50 POINTS AND BRAINLIEST THANK YOUUU

Chemistry
2 answers:
Vadim26 [7]3 years ago
7 0
Iserge mo huhuhuhu nasapatalastas
adoni [48]3 years ago
6 0

Answer:

chemical symbol= Cl ( chlorine)

electrons= 17

2p electrons = 6

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If 16.5g of c6h14o2 are reacted vwith .499 mol of o2. How many moles of co2 should be produced
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Answer:

moles of carbon dioxide produced are 410.9 mol.

Explanation:

Given data:

Mass of C₆H₁₄O₂ = 16.5 g

Moles of O₂ = 499 mol

Moles of CO₂ = ?

First of all we will write the balance chemical equation.

2C₆H₁₄O₂  +  17O₂  →   14CO₂  +  12H₂O

moles of C₆H₁₄O₂  = mass × molar mass

moles of C₆H₁₄O₂ =  16.5 g × 118 g/mol

moles of C₆H₁₄O₂ = 1947 mol

Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.

                 O₂      :     CO₂

                 17       :      14

                499     :      14/17× 499 = 410.9 moles

          C₆H₁₄O₂   :   CO₂

                    2    :      14

                 1947 :     14/2× 1947 =  13629 moles

Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.

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ankoles [38]

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Explanation:


May be you have experienced that: when you go to the beach, where the atmposhpere pressure is greater than the atmosphere pressure in places that are at higher altitudes, the water takes longer to boil. That is because the boiling temperature is greater, and you need more total heat (more time) to permit the liquid to reach that temperature.


The reason why that happens is because substances boil when the vapor pressure (the pressure of the particles of vapor over the liquid) equals the atmosphere pressure. So, when the atmposhere pressure increases, the temperature at which the vapor pressure reaches the atmosphere pressure also increases, and when the atmosphere pressure decreases, the temperature at which the vapor pressure reaches the atmosphere pressure decreases.



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A 0.0100 mol sample of Ca(OH)2 requires 50.00 mL of aqueous HCl for neutralization according to the reaction below. What is the
harkovskaia [24]

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