Answer:
4.78 mol O2
Explanation:
Grams --> moles
Grams/molar mass = moles
153 g O2 / 32 g/mol O2 = 4.78125 (significant figures so round it up; the grams has 3 significant figures, so round it up) 4.78 mol O2
Na2CO3 + 2Cl- ⇒ 2NaCl + CO3^-2
<span>
1 mole of Na2CO3 = 106 g </span>
<span>2 moles of NaCl = 2 x 58.4
= 116.8 g
</span>Na2CO3 would increase by 116.8 / 106 = 1.10 to form 2NaCl.
<span>0.4862 g x 1.10 = 0.515 grams of NaCl.
</span>
K2CO3 + 2Cl- ⇒ 2KCl + CO3^-2
<span>1 mole of K2CO3 = 138.2 g </span>
<span>2 moles of KCl = 149.1 </span>
<span>
K2CO3 would increase by </span>149.1 /138.2 = 1.079 <span>to form 2KCl
</span>
<span> 0.4862 x 1.079 = 0.5246 g</span>
Answer:
The Kc of this reaction is 311.97
Explanation:
Step 1: Data given
Kp = 0.174
Temperature = 243 °C
Step 2: The balanced equation
N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 3: Calculate Kc
Kp = Kc *(RT)^Δn
⇒ with Kp = 0.174
⇒ with Kc = TO BE DETERMINED
⇒ with R = the gas constant = 0.08206 Latm/Kmol
⇒ with T = the temperature = 243 °C = 516 K
⇒ with Δn = number of moles products - moles reactants 2 – (1 + 3) = -2
0.174 = Kc (0.08206*516)^-2
Kc = 311.97
The Kc of this reaction is 311.97
Answer:
The rate reduces by a factor cf /2.
