Which is one characteristic of producers?

If the theoretical yield of RX is 56.0 g , what is it’s percent yield ?b

Angelina_Jolie [31]

<span>the theoretical yield which is the expected yield and the actual yield obtained are not always the same. therefore percent yield is calculated which shows how much of the percentage of the theoretical yield is actually obtained.
the theoretical yield = 56.0 g
actual yield = 47.0 g
percent yield = actual yield / theoretical yield x 100 %
percent yield = 47.0 / 56.0 x 100% = 83.9 %
percent yield = 83.9 %</span>

8 0

4 years ago

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Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a cer

Sergeeva-Olga [200]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is $K_c= 2.8*10^{-4}$

Explanation:

From the question we are told that

The chemical reaction equation is

$Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}$

The voume of the misture is $V_m = 5.4L$

The molar mass of $Fe_{2} O_{3}_{(s)}$ is a constant with value of $M_{Fe_{2} O_{3}_{(s)} } = 160g/mol$

The molar mass of $H_{2}_{(g)}$ is a constant with value of $H_2 = 2g/mol$

The molar mass of $H_{2}O$ is a constant with value of $H_2O = 18g/mol$

Generally the number of moles is mathematically given as

$No \ of \ moles \ = \frac{mass}{molar\ mass}$

For $Fe_{2} O_{3}_{(s)}$

$No \ of\ moles = \frac{3.54}{160}$

$= 0.022125 \ mols$

For $H_{2}$

$No \ of\ moles = \frac{3.63}{2}$

$= 1.815 \ mols$

For $H_{2}O$

$No \ of\ moles = \frac{2.13}{18}$

$= 0.12 \ mols$

Generally the concentration of a compound is mathematicallyrepresented as

$Concentration = \frac{No \ of \ moles }{Volume }$

For $Fe_{2} O_{3}_{(s)}$

$Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}$

$= 4.10*10^{-3}M$

For $H_{2}$

$Concentration[H_2] = \frac{1.815}{5.4}$

$= 0.336M$

For $H_{2}O$

$Concentration [H_2O] = \frac{0.12}{5.4}$

$= 0.022M$

The equilibrium constant is mathematically represented as

$K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}$

Considering $H_2O \ for \ product$

And $H_2 \ for \ reactant$

At equilibrium the

$K_c = \frac{0.022}{0.336}$

$K_c= 2.8*10^{-4}$

3 0

3 years ago

If 9.6 X 1021 molecules of hydrogen are reacted with excess nitrogen, how many liters of ammonia can be produced at STP?

diamong [38]

The volume of NH₃ produced at STP : 0.237 L

<h3>Further explanation</h3>

Reaction

N₂ + 3H₂ → 2NH₃

1 mol = 6.02 x 10²³ particles

9.6 X 10²¹ molecules of Hydrogen, mol :

$\tt \dfrac{9.61\times 10^{21}}{6.02\times 10^{23}}=1.596\times 10^{-2}~moles$

mol H₂ : mol NH₃ = 3 : 2

mol NH₃ :

$\tt \dfrac{2}{3}\times 1.596\times 10^{-2}=0.0106$

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). <em>At STP, Vm is 22.4 liters/mol.</em>

The volume of NH₃ :

$\tt 0.0106\times 22.4=0.237~L$

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8 0

3 years ago

Which type of power can be used to reduce the production of carbon dioxide?

klasskru [66]

A. solar. Hope it helped!

8 0

3 years ago

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A 70.0-g piece of copper metal at 86.0 ∘c is placed in 50.0 g of water at 16.0 ∘c. The metal and water come to the same temperat

valentina_108 [34]

Copper is current voltage above 40 or 200 . copper useful for motor and some medicine

4 0

3 years ago