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katrin [286]
3 years ago
11

What is the mass ratio and atomic ratio for cl2o7

Chemistry
1 answer:
elixir [45]3 years ago
6 0

Answer:

Mass ratio: 71:112

Atomic ratio: 2:7

Explanation:

1. To get the mass ratio of of the compound Cl2O7, the following steps are followed:

- The atomic mass of Cl = 35.5g/mol, O = 16g/mol

- The mass of each element in the compound is as follows:

Cl2 = 35.5(2) = 71g

O7 = 16(7) = 112g

The mass ratio is the ratio of one mass of an element to another in the compound, hence, the mass ratio is 71:112

- The molar mass of the compound, Cl2O7, is determined:

71 + 112 = 183g/mol

- The mass percent of each element is determined by dividing the mass of each element present by the Molar Mass:

Cl = 71/183 = 0.3879 × 100 = 38.8%

O = 112/183 = 0.612 × 100 = 61.2%

2. Atomic ratio is the ratio of one atom in a molecule to another. It can be calculated this:

In Cl2O7, there are 9 total atoms (2 atoms of Cl + 7 atoms of oxygen).

Hence, that atomic ratio of Cl to oxygen in Cl2O7 is 2:7

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Answer:

m_{H_2O}=39.0g

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})

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m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g

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Generally, what happens to the rate of dissolution for a solid solute when you change from stirring a solution to not stirring i
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Which equation best represents the net ionic equation for the reaction that occurs when aqueous solutions of potassium phosphate
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2PO₄³⁻ + 3Fe²⁺ → Fe₃(PO₄)₂(s)

Explanation:

In a net ionic equation you list <em>only the ions that are participating in the reaction. </em>

When potassium phosphate, K₃PO₄, reacts with iron (II) nitrate, Fe(NO₃)₂ producing iron (II) phosphate, Fe₃(PO₄)₂ that is an insoluble salt. The reaction is:

2K₃PO₄ + 3 Fe(NO₃)₂ → Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺

The ionic equation is:

6K⁺ + 2PO₄³⁻ + 3Fe²⁺ + 6NO₃⁻→ Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺

Subtracting the K⁺ and NO₃⁻ ions that are not participating in the reaction, the net ionic equation is:

<h3>2PO₄³⁻ + 3Fe²⁺ → Fe₃(PO₄)₂(s)</h3>

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Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c
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Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓  +  Mg(NO₃)₂(aq)

This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:

           PbCl₂(s)  ⇄  Pb²⁺ (aq)  +  2Cl⁻ (aq)     Kps

Initial        x

React       s

Eq          x - s              s                  2s

As this is an equilibrium, the Kps works as the constant (Solubility product):

Kps = s . (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

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