Answer:
2 moles
Explanation:
Let us first start by calculating the molecular mass of Al₂O₃.
The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.
Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.
Molecular mass of Al : 26.981539 u
Molecular mass of O: 15.999 u
Therefore, molecular mass of Al₂O₃ is:
= 
u
= 101.960078 u
This can be approximated to 102 u.
1mole weighs 102 u
So, 2moles will weigh 2*102 = 204 u
How many protons does Thorium have? 90
How many neutrons does Thorium-234 have? 144
Calculate the mass defect for the isotope thorium-234 1.85864 amu
Answer:
hope it helps brainliest pls
Explanation:
Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54
<u>Answer:</u> The equilibrium concentration of water is 0.597 M
<u>Explanation:</u>
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For a general chemical reaction:
The expression for 
is written as:
![K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
The concentration of pure solids and pure liquids are taken as 1 in the expression.
For the given chemical reaction:
The expression of 
for above equation is:
![K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2S%5D%5E2%5Ctimes%20%5BO_2%5D%7D)
We are given:
![[H_2S]_{eq}=0.671M](https://tex.z-dn.net/?f=%5BH_2S%5D_%7Beq%7D%3D0.671M)
![[O_2]_{eq}=0.587M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D0.587M)
Putting values in above expression, we get:
![1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}](https://tex.z-dn.net/?f=1.35%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%280.671%29%5E2%5Ctimes%200.587%7D)
![[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%5Csqrt%7B%281.35%5Ctimes%200.671%5Ctimes%200.671%5Ctimes%200.587%29%7D%3D0.597M)
Hence, the equilibrium concentration of water is 0.597 M
