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3 years ago

How many moles of carbon dioxide are present in 14g of co2

Chemistry

1 answer:

Tomtit [17]3 years ago

8 0

0.32 moles

You just do 14/ 44.01 (the atomic mass of carbon + oxygen's atomic massX2)

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The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to conve

coldgirl [10]

Answer:

80.27%

Explanation:

Let's consider the following balanced equation.

2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)

First, we have to calculate the moles of Sn²⁺ that react.

$\frac{0.1015molSn^{2+} }{1L} .13.28 \times 10^{-3} L=1.348\times 10^{-3}molSn^{2+}$

We also know the following relations:

According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
The molar mass of Fe is 55.84 g/mol.

Then, for 1.348 × 10⁻3 moles of Sn²⁺:

$1.348\times 10^{-3}molSn^{2+}.\frac{2molFe^{3+} }{1molSn^{2+} } .\frac{1molFe}{1molFe^{3+} } .\frac{55.84gFe}{1molFe} =0.1505gFe$

If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:

$\frac{0.1505g}{0.1875g} \times 100 \% = 80.27\%$

5 0

3 years ago

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

6 0

3 years ago

How many moles are 21.67 L of NH4CI? Type your answer...

NNADVOKAT [17]

Answer:

0.967mole

Explanation:

Given parameters:

Volume of NH₄Cl = 21.67L

Unknown:

Number of moles = ?

Solution:

We assume that the volume was taken at standard temperature and pressure,

Then;

Number of moles = $\frac{volume }{22.4L}$

Number of moles = $\frac{21.67}{22.4}$ = 0.967mole

6 0

3 years ago

What is the name of this element?

Strike441 [17]

Answer:

Tungsten

Explanation:

8 0

3 years ago

Invasive Species Project!!

bonufazy [111]

Refer to picture below:

Kavinsky

4 0

2 years ago

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