calcium,phosphorus,potassium,and sulfer
10. You demonstrated the difference in density of the two objects. It is a physical property.
11. First calculate the density for all of them: density = mass/volume
Density:
A. 5/6 g/ml
B. 10/9 g/ml
C. 15/16 g/ml
D. 20/10 g/ml
If the density of the substance is higher than the density of the substance it is put in, then it will sink. So substances B and D will sink in water, as their densities are higher than 1 g/ml.
12. Ammonia weighs less than water does-- for example, the weight of 8 gallons of ammonia will be equivalent to the weight of 5 gallons of water.
Hope this helped!
Answer:
lithium
Explanation:
this is because lithium has a valency of 1 and oxygen has a valency of 2 thereby exchanging valency to create Li²0
Answer:-
Oxygen gains electrons and is reduced.
Explanation:-
For this reaction the balanced chemical equation is
4Fe + 3O2 --> 2Fe2O3
When Oxygen is present as oxygen gas, the oxidation number of O is Zero since it is the only element present in Oxygen gas.
Similarly Iron is present in Fe with oxidation number Zero.
In the case of Fe2O3, Oxygen has the oxidation number -2 while Iron has +3.
So the oxidation number of Oxygen goes from Zero to -2.
Since the oxidation number decreases Oxygen is reduced.
Since reduction involves gain of electrons, Oxygen gains electrons.
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
