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2 years ago

7

Fe (s) + S (I) --> Fes (s) In the experiment above, 7.62 g of Fe

1 answer:

Natali [406]2 years ago

6 0

Answer:

B

Explanation:

calculation

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¿Qué son los lantánidos y actínidos?

harkovskaia [24]

Answer: En estos grupos de los elementos de transición se encuentran las llamadas tierras raras, separadas del resto de elementos de la tabla, que pertenecen al grupo IIIB y se les conoce como lantánidos y actínidos.

6 0

2 years ago

______ is a chemical property of matter.

Reika [66]

I think it’s chemical reactivity

4 0

2 years ago

What is the relationship between metals and the number of electrons in the outer level?

gayaneshka [121]

Elements in group 1-2, 13-18, the number of valence electrons is related to the group number. For example, in the first group, the alkali metals there is one valence electron, however in group 13, there are 3 valence electrons. Valence electrons are also known as the outershell electrons.

4 0

2 years ago

Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What

Mariana [72]

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $CH_4$ :-

Mass of $CH_4$ = 1.28 g

Molar mass of $CH_4$ = 16.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.28\ g}{16.04\ g/mol}$

$Moles_{CH_4}= 0.0798\ mol$

For $O_2$ :-

Mass of $O_2$ = 10.1 g

Molar mass of $O_2$ = 31.998 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{10.1\ g}{31.998\ g/mol}$

$Moles_{O_2}= 0.3156\ mol$

According to the given reaction:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

<u>Limiting reagent is the one which is present in small amount. Thus, $CH_4$ is limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0798\ moles= \frac{Mass}{44.01\ g/mol}$

Mass of $CO_2$ = 3.51 g

<u> Theoretical yield = 3.51 g</u>

3 0

2 years ago

In this assignment, you will write and balance a chemical equation. Then, you will use a table of enthalpy values to calculate t

olganol [36]

Answer:

2 H2(g) + O2(g) → 2 H2O(ℓ) ΔH = −570 kJ

Explanation:

8 0

2 years ago