Zn = 28.15%
Cl = 30.53%
O = 41.32%
<h3>Further explanation</h3>
Given
Zn(CIO3)2 compound
Required
The % composition
Solution
Ar Zn = 65.38
Ar Cl = 35,453
Ar O = 15,999
MW Zn(CIO3)2 = 232.3
Zn = 65,38/232.3 x 100% = 28.15%
Cl = (2 x 35.453) / 232.3 x 100% = 30.53%
O = (6 x 15.999) / 232.3 x 100% = 41.32%
The volume of a sample of ammonia gas : 5.152 L
<h3>Further explanation</h3>
Given
0.23 moles of ammonia
Required
The volume of a sample
Solution
Assumed on STP
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So for 0.23 moles :
= 0.23 x 22.4 L
= 5.152 L
Answer:
The answer is B. False
Explanation:
The ratio of sizes between the ionic radii of cations and anions in a cell influences the manner of packing for that cell thereby predicting the possible cation/anion coordination number in any compound and establishing the structure of ionic solids.
Answer:
P = 1/8
Explanation:
The wave function of a particle in a one-dimensional box is given by:
Hence, the probability of finding the particle in the one-dimensional box is:
Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:
![P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]](https://tex.z-dn.net/?f=%20P%20%3D%20%5Cfrac%7B2%7D%7BL%7D%20%5B%5Cfrac%7BL%7D%7B16%7D%20%281%20-%204%5Cfrac%7Bsin%28%5Cfrac%7Bn%20%5Cpi%7D%7B4%7D%29%7D%7Bn%20%5Cpi%7D%29%5D%20)
Solving for n=4:
I hope it helps you!
Answer:
16.933g approximately 17.0 grams
Explanation:
From the simple promotions and given the same compound ascorbic acid (vitamin C)
In the laboratory synthesised ascorbic acid
Mass of carbon = 30.0g
Mas of Oxygen = 40.0g
That is the mass of Oxygen per unit mass of Carbon
Per gram of Carbon we have
(30.0g Carbon)÷30 combines with (40.0g of Oxygen)÷30
That is 4/3g of Oxygen per gram of Carbon
Hence the mass of Oxygen compound that combines with 12.7g of Carbonin natural occurring ascorbic acid (vitamin C) is = 4/3×12.7 = 16.933g approximately 17.0g
