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2 years ago

How many grams of FeCo3 will be produced from 57.2g FeCl2

Chemistry

1 answer:

Evgesh-ka [11]2 years ago

4 0

Answer:

287.30 g of FeCO₃

Solution:

The Balance Chemical Equation is as follow,

FeCl₂ + Na₂CO₃ → FeCO₃ + 2 NaCl

Step 1: Calculate Mass of FeCl₂ as,

Molarity = Moles ÷ Volume

Solving for Moles,

Moles = Molarity × Volume

Putting Values,

Moles = 2 mol.L⁻¹ × 1.24 L

Moles = 2.48 mol

Also,

Moles = Mass ÷ M.Mass

Solving for Mass,

Mass = Moles × M.Mass

Putting Values,

Mass = 2.48 mol × 126.75 g.mol⁻¹

Mass = 314.34 g of FeCl₂

Step 2: Calculate Mass of FeCO₃ formed as,

According to equation,

126.75 g (1 mole) FeCl₂ produces = 115.85 g (1 mole) FeCO₃

So,

314.34 g of FeCl₂ will produce = X g of FeCO₃

Solving for X,

X = (314.34 g × 115.85 g) ÷ 126.75 g

X = 287.30 g of FeCO₃

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A hypothetical element consists of four isotopes. Use the information below to calculate the average atomic volume mass of the e

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Answer:

104.969 amu.

Explanation:

From the question given above, the following data were obtained:

Isotope A:

Mass of A = 107.977 amu

Abundance (A%) = 0.1620%

Isotope B:

Mass of B = 106.976 amu

Abundance (B%) = 1.568%

Isotope C:

Mass of C = 105.974 amu

Abundance (C%) = 47.14%

Isotope D:

Mass of D = 103.973 amu

Abundance (D%) = 51.13%

Average atomic mass =?

The average atomic mass of the element can be obtained as follow:

Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]

Average atomic mass = [(107.977 × 0.1620)/100] + [(106.976 × 1.568)/100] + [(105.974 × 47.14)/100] + [(103.973 × 51.13)/100]

= 0.175 + 1.677 + 49.956 + 53.161

= 104.969 amu

Therefore, the average atomic mass of the element is 104.969 amu.

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3 years ago

Carboxylic acids and alcohols react via dehydration and condensation to produce a/an _______ and water.

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Answer:

ester

Explanation:

Carboxylic acids and alcohols react in the presence of strong acid to produce an ester and water. The two carbon chains come together, which is the condensation, and then water is generated, which is dehydration. The resulting compound is an ester.

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2 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

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The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

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Answer:

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