Answer:
a. 0.5 mol
b. 1.5 mol
c. 0.67
Explanation:
Fe3+ + SCN- -----> [FeSCN]2+
a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium
b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium
c. KC = 0.5/(0.5*1.5) = 0.67
n = 1.5atm (15L) / .0821 (280k) = .98 mol NaCl
NaCl = 22.99g Na + 35.45g Cl = 58.44g NaCl
58.44g NaCl x .98 mol NaCl = 57.27g NaCl
Explanation:
hope you get it right :)
Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
Answer: Since one molecule of CO2 contains one molecule carbon atoms, therefore, 0.6 molecule of carbon dioxide will contain 0.6 molecule of carbon.
Mass of C in 0.6 molecule of CO2
Explanation
= No. Of molecule × molar mass
= 0.6× 12gm
=7.2gm
<u>Volume </u><u>of</u><u> 106.9 mL from the concentrated solution should be taken and diluted to 350 </u><u>mL.</u>
⠀
⠀
<u>Main I'd On Indian Brainly Is - HeartCrush</u>
<u>We can use the </u><u>formula.</u><u> </u>
c1v1 =c2v2
<u>Where c1 is the concentration and v1 is volume of the concentrated </u><u>solution.</u><u> </u>
c2 is the concentration
and v2 is the volume of the diluted solution to be prepared
9.00 M x V1 = 2.75 M x 350 mL
V1 = 106.9 mL
<u>Volume of 106.9 mL from the concentrated solution should be taken and diluted to 350 </u><u>mL</u><u>.</u>
