Answer:
C. 1.35
Explanation:
2NH3 (g) <--> N2 (g) + 3H2 (g)
Initial concentration 2.2 mol/0.95L 1.1 mol/0.95L 0
change in concentration 2x x 3x
-0.84 M +0.42M +1.26M
Equilibrium 1.4 mol/0.95L=1.47M 1.58 M 1.26 M
concentration
Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M
Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M
Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M
K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M
Answer:
Steak is abotic living organisms are abotic and steak was abiotic
Explanation:
<span> </span>
Answer
is: volume is 20 mL.<span>
c</span>₁(CH₃COOH) = 2,5 M.<span>
c</span>₂(CH₃COOH) = 0,5 M.<span>
V</span>₂(CH₃COOH) = 100 mL.<span>
V</span>₁(CH₃COOH) = ?<span>
c</span>₁(CH₃COOH) · V₁(CH₃COOH)
= c₂(CH₃COOH) · V₂(CH₃COOH).<span>
2,5 M · V</span>₁(CH₃COOH)
= 0,5 M · 100 mL.<span>
V</span>₁(CH₃COOH) = 0,5 M · 100 mL ÷ 2,5 M.<span>
V</span>₁(CH₃COOH) = 20 mL ÷ 1000 mL/L =0,02 L.
Answer:
3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate.
Explanation:
![Ba(OH)_2.8H_2O(s)+NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+8H_2O(l)+NH_3(g)](https://tex.z-dn.net/?f=Ba%28OH%29_2.8H_2O%28s%29%2BNH_4SCN%28s%29%5Crightarrow%20Ba%28SCN%29_2%28s%29%2B8H_2O%28l%29%2BNH_3%28g%29)
The balance chemical equation is :
![Ba(OH)_2.8H_2O(s)+2NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+10H_2O(l)+2NH_3(g)](https://tex.z-dn.net/?f=Ba%28OH%29_2.8H_2O%28s%29%2B2NH_4SCN%28s%29%5Crightarrow%20Ba%28SCN%29_2%28s%29%2B10H_2O%28l%29%2B2NH_3%28g%29)
Mass of barium hydroxide octahydrate = 6.5 g
Moles of barium hydroxide octahydrate = ![\frac{6.5 g}{315 g/mol}=0.020635 mol](https://tex.z-dn.net/?f=%5Cfrac%7B6.5%20g%7D%7B315%20g%2Fmol%7D%3D0.020635%20mol)
According to reaction, 2 moles of ammonium thiocyanate reacts with1 mole of barium hydroxide octahydrate. The 0.020635 moles of barium hydroxide octahydrate will react with:
![\frac{2}{1}\times 0.020635 mol=0.04127 mol](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B1%7D%5Ctimes%200.020635%20mol%3D0.04127%20mol)
Mass of 0.04127 moles of ammonium thiocyanate;
![0.04127 mol\times 76 g/mol=3.136 g\approx 3.14 g](https://tex.z-dn.net/?f=0.04127%20mol%5Ctimes%2076%20g%2Fmol%3D3.136%20g%5Capprox%203.14%20g)
3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate