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4 years ago

Which energy profile best shows that the enthalpy of formation of cs2 is 89.4 kj/mol?

1 answer:

5 0

Answer is: B. C(s) + 2S(s) + 89.4 kJ → CS2(l).

Missing question:
A. C(s) + 2S(s) → CS2(l) + 89.4 kJ.
B. C(s) + 2S(s) + 89.4 kJ → CS2(l).
C. C(s) + 2S(s) + 89.4 kJ → CS2(l) + 89.4 kJ.
D. C(s) + 2S(s) → CS2(l).
Because enthalpy of the system is greater that zero, this is endothermic reaction (<span>chemical reaction that absorbs more energy than it releases)</span>, heat is included as a reactant.

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How many moles are in 550 grams of kic

ExtremeBDS [4]

Answer:

3.089653833499255

Explanation:

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3 0

3 years ago

Calculate the standard enthalpy change of formation of CHA given that the standard enthalpies change of combustion of methane, g

Nostrana [21]

Answer : The standard enthalpy of formation of methane is, -74.8 kJ/mole

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $CH_4$ will be,

$C(s)+2H_2(g)\rightarrow CH_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.4kJ/mole$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.7kJ/mole$

Now we will reverse the reaction 1, multiply reaction 3 by 2 then adding all the equations, we get :

(1) $CO_2(g)+2H_2O(l)\rightarrow CH_4(g)+2O_2(g)$ $\Delta H_1=890kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.4kJ/mole$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.7kJ)=-571.4kJ/mole$

The expression for enthalpy of formation of $CH_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+890kJ/mole)+(-393.4kJ/mole)+(-571.4kJ/mole)$

$\Delta H=-74.8kJ/mole$

Therefore, the standard enthalpy of formation of methane is, -74.8 kJ/mole

3 0

3 years ago

What are 1,2 and 3 examples of the same periodic table

Serga [27]

Period

Group

Explanation:

On the periodic table the numbers 1,2 and 3 can represent either groups or periods.

The periodic table gives the arrangement of elements based on their atomic numbers

They are divided into vertical columns and horizontal rows
The vertical columns are the groups whereas the horizontal rows are the periods.
The vertical columns are made up of elements having the same number of electrons in their outermost shell.
The horizontal columns are made up of elements having the same number of shell.
On the periodic table, the groups are numbered 1 - 18
There are 7 periods on the table.

learn more:

Periodic table brainly.com/question/1971327

#learnwithBrainly

5 0

4 years ago

How many moles of sodium nitrate, NaNO3, do you need to make 22.4L of oxygen gas at STP? *also if you can help with the other qu

max2010maxim [7]

<h2>Question no.18 </h2><h2>Part 1:</h2><h2>Answer:</h2>

We need 2 moles of sodium nitrate NaNO3 fro the production of one mole of oxygen gas.

<h3>Explanation:</h3>

The balanced chemical equation is:

2NaNO3 →→→→ 2NaNO2 + O2.

From the balanced chemical equation, it is obvious that for the production of One molecule of oxygen two molecules of NaNO3 breakdown.

So 22.4 L is equal to one mole.

Hence for the production of one mole of oxygen gas two moles of sodium nitrate will be needed.

<h2>Part b.</h2><h2>Answer:</h2>

The grams of NaNO3 for the production of 23.98 L of oxygen gas is 181.8786.

<h3>Explanation:</h3>

From the balanced chemical equation is:

2NaNO3 →→→→ 2NaNO2 + O2.

For the production of one mole of oxygen we need two moles of NaNO3.

So for 22. 4 L of O2 = 2 moles of NaNO3.

For:

1 L = 2/22.4

23.98 L = 2/22.4 * 23.98 = 2.14 moles.

In one mole of NaNO3, there are 84.99 grams.

So in 2.14 moles:

Mass in grams = 2.14 * 84.99 = 181.8786 g

Hence the grams of NaNO3 for the production of 23.98 L of oxygen gas is 181.8786.

<h2>Question 19</h2><h2>Part a:</h2><h2>Answer:</h2>

<u>If 48.8 L of the oxygen gas is used in the reaction then 48.8 liters of the carbon mono oxide gas will produce.</u>

<h3>Explanation:</h3>

From the balanced chemical equation:

C6H6S + 6O2 →→→→ 6CO + 3H2O + SO3

In the production of the six carbon monoxide six oxygen molecules are used up.

It means the ratio is 1 : 1.

It means the the amount of CO produced will be equal to the amount of O2 used.

So for the production of 48.8 L of carbon mono oxide production, the 48.8 L of the oxygen gas will be needed.

<h2>Part b.</h2><h2>Answer:</h2>

The 0.005648 L of the CO will produce with the use of 0.005648 L of oxygen gas.

<h3>Explanation:</h3>

From the balanced chemical equation:

C6H6S + 6O2 →→→→ 6CO + 3H2O + SO3

In the production of the six carbon mono oxide six oxygen molecules are used up.

It means the ratio is 1 : 1.

It means the the amount of CO produced will be equal to the amount of O2 used.

One mole of any gas is equal to 22.4 L.

So for the production of 0.005648 L of carbon mono oxide production, the 0.005648 L of the oxygen gas will be needed.

<h2>Part C.</h2><h2>Answer:</h2>

The 2.98 L of the carbon monoxide will be produced if we use 2.98 liters of oxygen gas.

<h3>Explanation:</h3>

From the balanced chemical equation, we can predict the amount of reactants and products in the chemical equation.

From the balanced chemical equation:

C6H6S + 6O2 →→→→ 6CO + 3H2O + SO3

In the production of the six carbon mono oxide six oxygen molecules are used up.

It means the ratio is 1 : 1.

It means the the amount of CO produced will be equal to the amount of O2 used.

One mole of any gas is equal to 22.4 L.

Hence for use of 2.97 L of oxygen gas, the 2.98 L of the carbon monoxide will be produced

7 0

3 years ago

A NaOH solution is standardized using the monoprotic primary standard potassium hydrogen phthalate, KHP (204.22 g/mol.) If 0.698

kakasveta [241]

Answer:

0.07789 M is the sodium hydroxide concentration.

Explanation:

Mass of potassium hydrogen phthalate = 0.6986 g

Molar mass of potassium hydrogen phthalate = 204.22 g/mol

Moles of potassium hydrogen phthalate = $\frac{0.6986 g}{204.22 g/mol}=0.003421 mol$

$NaOH+KHP\rightarrow NaKP+H_2O$

According to reaction , 1 mole og KHp reactswith 1 mole of NaOH , then 0.003421 moles of KHp will react with :

$\frac{1}{1}\times 0.003421 mol=0.003421 mol$

Moles of NaOH = 0.003421 mole

Volume of NaOH solution = 43.92 ml = 0.04392 L ( 1 mL = 0.001L)

$Concentration=\frac{Moles}{Volume(L)}$

Concentration of NaOH :

$\frac{0.003421 mol}{0.04392 L}=0.07789 M$

0.07789 M is the sodium hydroxide concentration.

5 0

3 years ago