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LenaWriter [7]
3 years ago
7

For the chemical reaction CaI2+2AgNO3-> 2AgI+Ca(NO3)2 how many moles of silver iodide will be produced from 205g of calcium i

odide
Chemistry
1 answer:
Colt1911 [192]3 years ago
6 0

Answer:

Moles of silver iodide produced = 1.4 mol

Explanation:

Given data:

Mass of calcium iodide = 205 g

Moles of silver iodide produced = ?

Solution:

Chemical equation:

CaI₂ + 2AgNO₃     →      2AgI + Ca(NO₃)₂

Number of moles calcium iodide:

Number of moles = mass/ molar mass

Number of moles = 205 g/ 293.887 g/mol

Number of moles = 0.7 mol

Now we will compare the moles of calcium iodide with silver iodide.

                     CaI₂         :           AgI

                         1           :             2

                       0.7         :           2×0.7 = 1.4

Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.

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ZanzabumX [31]

Answer:

- To increase the temperature as it is a reactant in terms of its endothermicity.

- To remove it will enable more space for the reactant to favor its production.

- To add more reactant in order to increase its equilibrium concentration.

Explanation:

Hello,

The undergoing chemical reaction is:

2NH_3(g)\rightleftharpoons 3H_2(g)+N_2(g)

Thus, in order to intensify the amount of nitrogen as the chemical reaction is endothermic, considering the Le Chatelier's principle we state:

- To increase the temperature as it is a reactant in terms of its endothermicity.

- To remove it will enable more space for the reactant to favor its production.

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Explanation:

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3 years ago
A 3.28 L solution is prepared by dissolving 535 g CaCl2 in water. The molar mass of CaCl2 is 110.98 g. What is the morality of t
andrey2020 [161]

Answer:

              1.47 mol/L

Explanation:

Molarity is given as,

                           Molarity = Moles / Vol in L    ------- (1)

Moles of CaCl₂,

                           Moles = Mass / M.Mass

                           Moles = 535 g / 110.98 g/mol

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Now, putting values in eq. 1.

                           Molarity = 4.82 mol / 3.28 L

                           Molarity = 1.47 mol/L

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