Answer:
Mass = 114.26 g
Explanation:
Given data:
Number of gold atoms = 3.47×10²³ atoms
Mass in gram = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
3.47×10²³ atoms × 1 mol /6.022 × 10²³ atoms
0.58 mol
Mass of gold:
Mass = number of moles × molar mass
Mass = 0.58 mol × 197 g/mol
Mass = 114.26 g
Answer:
CO is considered as a product.
Explanation:
A general chemical equation for a combination reaction follows:
To write a chemical equation, we must follow some of the rules:
The reactants must be written on the left side of the direction arrow.
A '+' sign is written between the reactants, when more than one reactants are present.
An arrow is added after all the reactants are written in the direction where reaction is taking place. Here, the reaction is taking place in forward direction.
The products must be written on the right side of the direction arrow.
A '+' sign is written between the products, when more than one products are present.
For the given chemical equation:
are the reactants in the reaction and are the products in the reaction.
Hence, CO is considered as a product.
Answer:
Option C.
Ilegal dumping of waste
Explanation:
This is because non point source of pollution refers to source of pollution that are many and not directly one which is illegal or does not meet the legal term. This type of pollution does not have a point source, it has many sources and this type of pollution is cause by rainfall or precipitation. Where when the rain fall, it wash away the waste through to water bodies, causing pollution and endangering water bodies.
Answer: aging?
Explanation: sorry, i’m not too sure, but that would be my best guess.
Answer: The enthalpy change for formation of butane is -125 kJ/mol
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B4%5Ctimes%20H_f_%7BCO_2%7D%2B5%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%20H_f_%7BO_2%7D%5D)
![-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]](https://tex.z-dn.net/?f=-2877%3D%5B%284%5Ctimes%20-393%29%2B%285%5Ctimes%20-286%29%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%200%5D)
Thus enthalpy change for formation of butane is -125 kJ/mol
