Answer:
pOH= 14.248
[H+]=1.77 M
[OH-]=5.65 x10^-15M
Explanation:
pH+pOH= 14
pOH= 14-pH
pOH=14-(-0.248)
pOH= 14.248
[H+]=10^-pH= 10^-(-0.248)=1.77 M
[OH-]=10^-pOH= 10^-14.248=5.65 x10^-15M
Yes mixing salt with pepper change
<h3>
Answer:</h3>
8 alpha particles
4 beta particles
<h3>
Explanation:</h3>
<u>We are given;</u>
- Neptunium-237
- Thallium-205
- Neptunium-237 undergoes beta and alpha decay to form Thallium-205.
We are required to determine the number of beta and alpha particles produced to complete the decay series.
- We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
- When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.
In this case;
Neptunium-237 has an atomic number 93, while,
Thallium-205 has an atomic number 81.
Therefore;
²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti
We can get x and y
237 = 4x + y(0) + 205
237-205 = 4x
4x = 32
x = 8
On the other hand;
93 = 2x + (-y) + 81
but x = 8
93 = 16 -y + 81
y = 4
Therefore, the complete decay equation is;
²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti
Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.
Answer:
C = 0.08M
Explanation:
molar mass of AlCl3
Al =27
Cl = 35.5
27+3(35.5) =133.5g/mol
n= mass/Molar mass
n =CV
CV = mass/molar mass
C x 500 x 10^-³ = 5/133.5
C x 500 x 10^-³ = 0.04
C = 0.04/500 x 10^-³
C = 0.08M
You can establish a system of two equation with two variables.
Varibles are:
V1 = volume of the 50% sugar solution
V2 = volumen of the 80% sugar solution
Equations:
Balance of sugar:
Sugar from 50% solution: 0.5*V1
Sugar from 80% solution: 0.8*V2
Sugar in the final solution (mix): 0.6 * 105 = 63
1) 0.5V1 + 0.8V2 = 63
Final volume = volume of 50% solution + volume of 80% solution
2) V1 + V2 = 105
From (2) V1 = 105 - V2
Substitue in (1)
0.5 (105 - V2) + 0.8 V2 = 63
52.5 - 0.5V2 + 0.8V2 = 63
0.3 V2 = 63 - 52.5
0.3 V2 = 10.5
V2 = 10.5/0.3
V2 = 35mL
V1 = 105 - 35 = 70 mL
Answer: 70 mL of the 50% solution and 35 mL of the 80% solution.
