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xz_007 [3.2K]
3 years ago
12

PLEASE HELP I HAVE LIMITED TIME!!

Chemistry
1 answer:
n200080 [17]3 years ago
3 0

Answer:

Explanation:

Si tomamos en cuenta el peso molecular del agua, que es equivalente a:

1 Átomo de H₂O

O = 16 gr/mol

H = 1 gr/mol

H₂O = 18 gr/mol

Teóricamente sabemos que en 1 mol de H2O habrá 18 gr.

 

Para obtener los moles presentes en 1 mg de H₂O, (como 1 gr = 1000 mg), decimos:

1 mol H2O …………………………..  18000 mg

       X          ……………………………   1 mg

X = 1 / 18000 = 5,56 X 10⁻⁵ moles de H20

 

Y para obtener la cantidad de moléculas presentes, de acuerdo a los moles, multiplicamos por el número de Avogadro (6,023 X 10²³ moléculas /mol)

Moléculas de H₂O = 5,56x 10⁻⁵ mol x 6,023 x 10²³

Moléculas de H₂O = 3,34488 x10¹⁹ moléculas de H₂O

En el copo de nieve habrá 3,34488 x 10¹⁹ moléculas de H₂O.

Espero que te sirva =)

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AlekseyPX
When CH₄ is burnt in excess O₂ following products are formed,

                           CH₄  +  2 O₂     →     CO₂  +  2 H₂O

According to equation 1 mole of CH₄ (16 g) reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O. Hence the products are,

                          1 mole of CO₂  and  2 moles of H₂O

Converting 1 mole CO₂ to grams;
As,
                           Mass  =  Moles × M.mass

                           Mass  =  1 mol ×  44 g.mol⁻¹

                           Mass  =  40 g of CO₂

Converting 2 moles of H₂O to grams,

                           Mass  =  2 mol ×  18 g.mol⁻¹
                         
                           Mass  =  36 g of H₂O

Total grams of products;

                           Mass of CO₂  =  44 g
                    +     Mass of H₂O  =  36 g
                                                  -------------
                           Total              =   80 g of Product

Result:
            80 grams of product
is formed when 16 grams of CH₄ is burnt in excess of Oxygen.
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3 years ago
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Answer:

238,485 Joules

Explanation:

The amount of energy required is a summation of heat of fusion, capacity and vaporization.

Q = mLf + mC∆T + mLv = m(Lf + C∆T + Lv)

m (mass of water) = 75 g

Lf (specific latent heat of fusion of water) = 336 J/g

C (specific heat capacity of water) = 4.2 J/g°C

∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C

Lv (specific latent heat of vaporization of water) = 2,260 J/g

Q = 75(336 + 4.2×139 + 2260) = 75(336 + 583.8 + 2260) = 75(3179.8) = 238,485 J

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Answer:

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Explanation:

Based on the equation:

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<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>

<em />

To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:

<em>Moles Cl₂:</em>

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<em>Moles KI -Molar mass: 166.0028g/mol-</em>

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Here, clarely, the KI is the limiting reactant

As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:

0.15 moles * (74.5513g / mol) =

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Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100

<h3>Percent yield = 89.1%</h3>
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