Answer: The balanced equation is ![2FeCl_{3} + 3MgO \rightarrow Fe_{2}O_{3} + 3MgCl_{2}](https://tex.z-dn.net/?f=2FeCl_%7B3%7D%20%2B%203MgO%20%5Crightarrow%20Fe_%7B2%7DO_%7B3%7D%20%2B%203MgCl_%7B2%7D)
Explanation:
The given reaction equation is as follows.
![FeCl_{3} + MgO \rightarrow Fe_{2}O_{3} + MgCl_{2}](https://tex.z-dn.net/?f=FeCl_%7B3%7D%20%2B%20MgO%20%5Crightarrow%20Fe_%7B2%7DO_%7B3%7D%20%2B%20MgCl_%7B2%7D)
Here, number of atoms present on reactant side are as follows.
Number of atoms on product side are as follows.
To balance this equation, multiply
by 2 and MgO by 3 on reactant side. Also, multiply
by 3 on product side. Therefore, the equation can be rewritten as follows.
Hence, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.
Thus, we can conclude that the balanced equation is
.
Answer:
4
10
Explanation:
The reaction equation is given as;
Ca(OH)₂ → Ca²⁺ + 2OH⁻
Concentration of Ca(OH)₂ = 5 x 10⁻⁵M
Unknown:
pOH of the solution = ?
pH of the solution = ?
Solution:
Solve for the pOH of this solution using the expression below obtained from the ionic product of water;
pOH = ⁻log₁₀[OH⁻]
Ca(OH)₂ → Ca²⁺ + 2OH⁻
1moldm⁻³ 1moldm⁻³ 2 x 1moldm⁻³
5 x 10⁻⁵moldm⁻³ 5 x 10⁻⁵moldm⁻³ 2( 5 x 10⁻⁵moldm⁻³ )
1 x 10⁻⁴moldm⁻³
Therefore;
pOH = -log₁₀ 1 x 10⁻⁴ = 4
Since
pOH + pH = 14
pH = 14 - 4 = 10
180 degrees
if it is bent (meaning there is a lone pair of electrons on Se) it is 109.5 degrees
Answer: The volume of 9.7 tons of given ethylene glycol is 0.0079
.
Explanation:
Given: Density = 1.11 ![g/cm^{3}](https://tex.z-dn.net/?f=g%2Fcm%5E%7B3%7D)
Mass = 9.7 tons
Convert tons into lb as follows.
![1 ton = 2 \times 10^{3} lb\\9.7 ton = 9.7 ton \times 2 \times 10^{3} \frac{lb}{1 ton}\\= 19.4 \times 10^{3} lb](https://tex.z-dn.net/?f=1%20ton%20%3D%202%20%5Ctimes%2010%5E%7B3%7D%20lb%5C%5C9.7%20ton%20%3D%209.7%20ton%20%5Ctimes%202%20%5Ctimes%2010%5E%7B3%7D%20%5Cfrac%7Blb%7D%7B1%20ton%7D%5C%5C%3D%2019.4%20%5Ctimes%2010%5E%7B3%7D%20lb)
Now, lb is converted into kg as follows.
![1 kg = 2.2046 lb\\1 lb = 0.453592 kg\\19.4 \times 10^{3} lb = 19.4 \times 10^{3} lb \times 0.453592 \frac{kg}{1 lb}\\= 8.79 kg](https://tex.z-dn.net/?f=1%20kg%20%3D%202.2046%20lb%5C%5C1%20lb%20%3D%200.453592%20kg%5C%5C19.4%20%5Ctimes%2010%5E%7B3%7D%20lb%20%3D%2019.4%20%5Ctimes%2010%5E%7B3%7D%20lb%20%5Ctimes%200.453592%20%5Cfrac%7Bkg%7D%7B1%20lb%7D%5C%5C%3D%208.79%20kg)
1 kg = 1000 g
So, 8.79 kg = 8790 g
Density is the mass of a substance divided by its volume. Hence, volume of ethylene glycol is calculated as follows.
![Density = \frac{mass}{volume}\\1.11 g/cm^{3} = \frac{8790 g}{volume}\\volume = 7918.92 cm^{3} (1 cm^{3} = 10^{-6} m^{3})\\= 0.0079 m^{3}](https://tex.z-dn.net/?f=Density%20%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%5C%5C1.11%20g%2Fcm%5E%7B3%7D%20%3D%20%5Cfrac%7B8790%20g%7D%7Bvolume%7D%5C%5Cvolume%20%3D%207918.92%20cm%5E%7B3%7D%20%281%20cm%5E%7B3%7D%20%3D%2010%5E%7B-6%7D%20m%5E%7B3%7D%29%5C%5C%3D%200.0079%20m%5E%7B3%7D)
Thus, we can conclude that the volume of 9.7 tons of given ethylene glycol is 0.0079
.