Answer:
0.5 × 10²³ atoms of iodine
Explanation:
Given data:
Mass of calcium iodide = 12.75 g
Number of atoms of iodine = ?
Solution:
First of all we will calculate the number of moles of calcium iodide.
Number of moles = mass/ molar mass
Number of moles = 12.75 g/ 293.9 g/mol
Number of moles = 0.04 mol
In one mole of calcium iodide there are two moles of iodine.
Thus in 0.04 moles:
0.04 mol × 2 = 0.08 moles of iodine
Now we will use the Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
0.08 moles of iodine × 6.022 × 10²³ atoms / 1 mol
0.5 × 10²³ atoms of iodine.
A 0.00143 M concentration of MnO4^- is not a reasonable solution .
<h3>Number of moles of carbonate</h3>
The ions left in solution are Na^+ and NO3^-
Number of moles of calcium nitrate = 100/1000 L × 1 = 0.1 moles
Since;
1 mole of sodium carbonate reacts with 1 mole of calcium nitrate then 0.1 moles of sodium carbonate were used.
<h3>Conductivity of filtrate</h3>
The claim of the student that the concentration of sodium carbonate is too low is wrong because the value was calculated from concentration and volume of calcium nitrate and not using the precipitate. If the filtrate is tested for conductivity, it will be found to conduct electricity because it contains sodium and NO3 ions.
2) In the reaction as shown, the MnO4^- ion was reduced.
The initial volume is 3.4 mL while the final volume is 29.6 mL.
Number of moles of MnO4^- ion = (29.6 mL - 3.4 mL)/1000 × 0.0235 M = 0.0006157 moles
<h3>The calculations are performed as follows</h3>
- If 2 moles of MnO4^- reacted with 5 moles of acid
0.0006157 moles of MnO4^- reacted with 0.0006157 moles × 5 moles/ 2 moles
= 0.0015 moles
- In this case, number of moles of acid = 0.139 g/90 g/mol = 0.0015 moles
Number of moles of MnO4^- = 0.00143 M × (29.6 mL - 3.4 mL)/1000
= 0.000037 moles
- If 2 moles of MnO4^- reacts with 5 moles of acid
0.000037 moles of MnO4^- reacts with 0.000037 moles × 5 moles/ 2 moles
= 0.000093 moles
- Hence, this is not a reasonable amount of solution.
Learn more about MnO4^- : brainly.com/question/10887629
Answer:
17.04 g/mol
Explanation:
Molar Mass of NH₃
we know that
Nitrogen has 14.01 gram/mol
And Hydrogen has 1.01 gram/mol
but we have 3 Hydrogens So we multiply
1.01 by 3 i.e., 3.03
Now, add
14.01
+<u> </u><u>3</u><u>.</u><u>0</u><u>3</u>
17.04
So, The molar mass of ammonia, NH₃ is
17.04 g/mol
<u>-TheUnknown</u><u>Scientist</u>
4.17 moles. Good luck! :)
Answer:
C. P = nRT
Explanation:
PV = nRT, where n is a number of moles and R is the universal gas constant, R = 8.31 J/mol ⋅ K.
Hope this helps :)
