It’s the third one because Cl has 17 protons bc of the numeric number and 18 electrons bc it’s always the opposite and 18 neutrons because you subtract 35-17=18
Answer:
Moles of silver iodide produced = 1.4 mol
Explanation:
Given data:
Mass of calcium iodide = 205 g
Moles of silver iodide produced = ?
Solution:
Chemical equation:
CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂
Number of moles calcium iodide:
Number of moles = mass/ molar mass
Number of moles = 205 g/ 293.887 g/mol
Number of moles = 0.7 mol
Now we will compare the moles of calcium iodide with silver iodide.
CaI₂ : AgI
1 : 2
0.7 : 2×0.7 = 1.4
Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.
Answer:
a) v = 1497.2 cm^-1
b) v = 1465 cm^-1
Explanation:
In the attached image is the procedure explained to reach the answer.
Answer:
phosphorous- 5
calcium- 2
nitrogen- 3 or 5
iron- 8 (transition metals use subshells as valence electrons)
argon- 8
potassium- 1
helium- 2
magnesium- 2
sulfur- 6
lithium- 1
iodine- 7
oxygen- 6
barium- 2
aluminum- 3
hydrogen- 1
xenon- 8
copper- 1
Source: my own chemistry notes
Answer:

Explanation:
In this case, we can start with the reaction:

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:


Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):


The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol
= 48 g
(therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

I hope it helps!