The atmospheric pressure will be:
The pressure of the atmosphere resulting from the mercury column is 0.959 atm
What is atmospheric pressure?
The force that an object experiences from the weight of the air above it per unit area are known as atmospheric pressure.
Given: Height of mercury column = 729 mm Hg
To find: The pressure of the atmosphere
Calculation:
The atmospheric column resulting from the mercury column is calculated as follows:
1 atm =760 mm Hg
So, we can convert the 729 mm Hg to atm, and we get
Atmospheric pressure = 729 x 1 atm / 760 = 0.959 atm
Learn more about atmospheric pressure here,
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The enthalpy of reaction or ∆H reaction is the difference between the bond energy of the reactants and the bond energy of the products.
<h3>What is ∆H reaction?</h3>
The term ∆H reaction refers to the heat that is evolved or absorbed in a chemical reaction. It is also known as the enthalpy of reaction.
The question is incoherent but I will try to answer as much as possible. Using the values of bond energy, ∆H reaction = Bond energy of reactants - bond energy of products. This will give us the enthalpy of reaction.
Learn more about bond energy: brainly.com/question/1657608
Answer:
The answer is
<h2>12,574,750.83 moles</h2>
Explanation:
In order to find the moles of CO2 we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question
N = 7.57 × 10^30 molecules of CO2
We have
We have the final answer as
<h3>12,574,750.83 moles</h3>
Hope this helps you
Answer:
The correct answer is: 6.6 g MgO
Explanation:
First we have to write and balance the chemical reaction as follows:
2Mg(s) + O₂(g) → 2MgO(s)
That means that 2 moles of Mg(s) react with 1 mol of O₂(g) to give 2 moles of MgO(s). If Mg is totally consumed and a mass of O₂ remains unaltered after reaction, t<em>he limiting reactant is Mg</em>. We use the limiting reactant to calculate the mass of product.
According to the balanced chemical equation, 2 moles of Mg(s) produce 2 moles of MgO(s).
2 moles Mg = 2 mol x molar mas Mg= 2 mol x 24.3 g/mol = 48.6 g Mg
2 moles MgO= 2 mol x (molar mass Mg + molar mass O) = 2 mol x (24.3 g/mol + 16 g/mol) = 80.6 g MgO
The stoichiometric ratio is 80.6 g MgO/48.6 g Mg. So, we multiply this ratio by the mass of consumed Mg (4.0 g) in order to obtain the produced mass of MgO:
4.0 g Mg x 80.6 g MgO/48.6 g Mg = 6.63 g MgO
6.6 grams of magnesium oxide are formed.
