Answer:
This question is incomplete, here's the complete question:
<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>
Explanation:
Reaction :-
K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4
Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol
Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol
Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L
Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L
Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol
Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.
0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+
Final concentration of potassium cation
= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M
The answer <span>is <span>8.9 g/mL</span>.</span>
The density (D) is <span>equal to mass (m) divided by volume (V): D = m/V
Let's find the mass of the object:
m = 156 g - 105.5 g = 50.5 g
Let's find the volume of the volume:
V = 30.7 mL - 25 mL = 5.7 mL
The density is:
D = m/V = 50.5 g / 5.7 mL = 8.9 g/mL</span>
That would be 3.621471•10^3
This element is found in group 3A, period 3
<h3>Further explanation
</h3>
The maximum number of electrons that can be filled in the nth electron shell is 2n²(n=shell)
-
K shell (n = 1) maximum 2 x 1² = 2 electrons
- L shell (n = 2) maximum 2 x 2² = 8 electrons
- M shell (n = 3) maximum 2 x 3² = 18 electrons
- N shell (n = 4) maximum 2 x 4² = 32 electrons
Electron configuration of element X : 2.8.3 , so :
K shell = 2 ⇒1s²
L shell = 8⇒2s²2p⁶
M shell = 3⇒ 3s²3p¹
Block p: group 13-18 (has a 2p-6p configuration), also called a representative element because it includes metals, non-metals and metalloids
The outer shell 3s²3p¹ : located in group 3A and period 3
group⇒valence electron ⇒3
period⇒the greatest value of the quantum number n⇒3
Answer:
from the 1st equation:
4NH3 4NO
4 *(68) 4*30
1216 X mass of NO = 536.5 g
from the 2nd Equation
2NO 2NO2
2*30 2* 46
536.5 x mass of NO2 = 822.6 grams
from the 3rd Equation
3NO2 2HNO3
3*(46) 2* (63)
822.6 X mass of nitric acid = 751.06 gram
b) % yields = ( 96.2%* 91.3% *91.4%)= 80.3%
