Answer:
atomic mass of X is 48.0 amu
Explanation:
Let y be the atomic mass of X
Molar mass of O_2 is = 2×16 = 32 g / mol
X + O2 -----> XO_2
According to the equation ,
y g of X reacts with 32 g of O_2
24 g of X reacts with Z g of O_2
Z = ( 32×24) / y
But given that 24.0 g of X exactly reacts with 16.0 g of O_2
So Z = 16.0
⇒ (32×24) / y = 16.0
⇒ y = (32×24) / 16
y= 48.0
So atomic mass of X is 48.0 amu
what do u need help with u pls respond quickly
1) The question contains an unknown unit
The number 8908.8 has to be in units of mass: for example, kg or grams.
Here you indicated L.
I am going to work assuming that L is a mass unit. So you can see the way to solve the problem, but you have to verifiy the real unit of the statement and substitute with it.
With that in mind you can find the density of the liquid from:
density = mass / volume
2) Calculate the volume.
The volume of the liquid is the volume of the vessel, because it is filled.
The volume of the vessel is calculated from the formula of volume for a rectantular prism.
Volume of a rectangular prism = area of the base * height = side * side * height
=> Volume = 10 dm * 4 dm * 6 dm = 240 dm^3 = 240 liter
3) Calculate the density:
density = mass /volume = 8,908.8 L / 240 liter = 37.12 L / dm^3
Answer: 37.12 L / dm^3
Answer:
B. Alpha
Explanation:
Alpha decays always split into an element and He.
Beta decays always split into an element and e- (eletrons).
Gamma decays always split into an element and radiation.
Answer:
a) [H₃O⁺] = 1.8x10⁻⁵ M
b) pH = 4.75
c) % rxn = 3.5x10⁻³ %
Explanation:
a) The dissociation reaction of HCN is:
HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)
0.5 M - x x x
The dissociation constant from the above reactions is given by:
![Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH_%7B3%7DO%5E%7B%2B%7D%5D%5BCN%5E%7B-%7D%5D%7D%7B%5BHCN%5D%7D%20%3D%206.17%20%5Ccdot%2010%5E%7B-10%7D)
By solving the above quadratic equation we have:
x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]
Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.
b) The pH is equal to:
![pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%20-log%281.75%20%5Ccdot%2010%5E%7B-5%7D%20M%29%20%3D%204.75)
Then, the pH of the HCN solution is 4.75.
c) The % reaction is the % ionization:
![\% = \frac{x}{[HCN]} \times 100](https://tex.z-dn.net/?f=%20%5C%25%20%3D%20%5Cfrac%7Bx%7D%7B%5BHCN%5D%7D%20%5Ctimes%20100%20)
Therefore, the % reaction or % ionization is 3.5x10⁻³ %.
I hope it helps you!
