What is the number of mole equivalents of H2 produced by the reaction of 1 mole equivalent of Mg (m) with H2SO4 (aq)?

Chemistry

1 answer:

Inga [223]3 years ago

8 0

Answer:

We'll have 1.00 mol H2

Explanation:

Step 1: Data given

Number of moles of Mg = 1.00 moles

Step 2: The balanced equation

Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

Step 3: Calculate moles of H2

For 1 mol Mg we need 1 mol H2SO4 to produce 1 mol MgSO4 and 1 mol H2

We'll have 1.00 mol H2

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How do you solve this ??

kotegsom [21]

Answer : Option A) 2.00 eV

Explanation : The conversion of J to eV is done with the following formula;

$E_{eV} = E_{J} X (6.241 X 10^{18})$

Here, we have the value of particle in terms of Joules which is 3.2 X $10^{-19}$

So, on substituting we get,
$E_{eV}$ = 3.2 X $10^{-19}$ X $(6.241 X 10^{18} )$

$E_{eV}$ = 1.99 eV so, it can be rounded off to 2.00 eV.

3 0

3 years ago

How many molecules are in 41.8 g of sulfuric acid

Anton [14]

Answer

× 10²³ molecules are in 41.8 g of sulfuric acid

Explanation

The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.

Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol

$Mole=\frac{Mass}{Molar\text{ }mass}=\frac{41.8\text{ }g}{98.079\text{ }g\text{/}mol}=0.426187053\text{ }mol$

Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.

Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.

Therefore, 0.426187053 moles of sulfuric acid is equal

$\frac{0.426187053\text{ }mol}{1\text{ }mol}\times6.022×10²³\text{ }molecules=2.57\times10^{23}\text{ }molecules$

Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.

3 0

1 year ago

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves

hjlf

This is an incomplete question, here is a complete question.

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide:

$2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)$

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

$3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)$

Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.

Answer : The net chemical equation for the production of manganese is:

$6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)$