Answer:
A. 
B. 
Explanation:
Hello!
In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:
Thus we proceed as follows:
A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:
Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:
B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:
Best regards!
Answer: It changed in identity and properties.
Explanation:
I believe the answer is c
Answer:
0,07448M of phosphate buffer
Explanation:
sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:
Na₂HP + HCl ⇄ NaH₂P + NaCl <em>(1)</em>
NaH₂P + HCl ⇄ H₃P + NaCl <em>(2)</em>
The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:
0,01862L of HCl×
= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.
The concentration is:
= <em>0,07448M of phosphate buffer</em>
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I hope it helps!
Answer:
Explanation:
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 58.12 44.01
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
m/g: 9.511
1. Moles of C₄H₁₀
2. Moles of CO₂
The molar ratio is 8 mol CO₂:2 mol C₄H₁₀
3. Mass of CO₂
