Answer:
You should start with 63.54 grams of copper.
Explanation:
The chemical reactions are processes in which the nature of the substances changes, that is, from some initial substances called reactants, totally different ones called products are obtained.
In the chemical reaction, the formulas of reagents and products appear preceded by numbers (the stoichiometric coefficients) that indicate the proportions according to which the transformation occurs. So you can say that stoichiometry establishes relationships between the molecules or elements that make up the reactants of a chemical equation with the products of said reaction. The relationships that are established are MOLAR relationships between the compounds or elements that make up the chemical equation: always in MOLES.
The stoichiometric coefficients of a chemical equation are due to the fact that the atoms present before the reaction must be the same after the reaction, although they will have been rearranged to produce new substances.
If you want 2 moles of silver (Ag), for stoichiometry of the reaction you need a moles of copper Cu. Being the molar mass of copper Cu 63.54 g / mole, then:
1 mole*63.54 g/mole= 63.54 g
<u><em>
You should start with 63.54 grams of copper.</em></u>
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The mass of chlorine required to react with 10.0 g of sodium metal to produce sodium chloride is 15.26 g.
The balanced chemical reaction is :
<em> 2Na + Cl₂ → 2NaCl</em>
<em />
Given data :
Mass of sodium = 10.0 g
Number of moles of Na -
Number of moles = <u> Given mass </u>
Molar mass
Number of moles = <u> 10.0 g</u> = 0.43 mol
23 g
Comparing moles of Na and Cl₂ from the balanced chemical equation,
Na : Cl₂
2 : 1
0.43 : <u> 1 </u> x 0.43 = 0.215 mol
2
Mass of chlorine gas :
(molar mass of chlorine = 71 g)
Mass of chlorine = number of moles x molar mass
Mass = 0.215 mol x 71 g/mol
= 15.26 g
To learn more about sodium and chlorine,
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Production of materials and transportation are the examples of three carbon emission.
Extraction and production of purchased materials and transportation of purchased fuels are the examples of three carbon emission. Scope 3 emissions refers to all indirect emissions that occur in the chain of the reporting company that is included in both upstream and downstream emissions.
Big machineries are used for the production and extraction of materials as well as the transportation requires fossil fuels for working which releases carbondioxide gas in the atmosphere so we can conclude that production of materials and transportation are the examples of three carbon emission.
Answer:
- 10.555 kJ/mol.
Explanation:
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).
∆H°rxn is the standard enthalpy change of the reaction (J/mol).
T is the temperature of the reaction (K).
∆S°rxn is the standard entorpy change of the reaction (J/mol.K).
∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants
<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.
∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants
<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>
<em></em>
∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.
<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>
Explanation:
Molar mass of O2: 32g/mol
Moles of O2 = 19.5 / 32 = 0.609mol
