Answer:
a. N₂O₅ + H₂O ⇒ 2 HNO₃ (pentóxido de dinitrógeno + agua ⇒ ácido nítrico)
b. Na₂O + H₂O ⇒ 2 NaOH (óxido de sodio + agua ⇒ hidróxido de sodio)
Explanation:
Tenemos que balancear, por el método de tanteo, las siguientes ecuaciones químicas.
a. En la primera reacción, el pentóxido de dinitrógeno reacciona con agua para formar ácido nítrico. Es una reacción de síntesis o combinación.
N₂O₅ + H₂O ⇒ HNO₃
Podremos obtener la ecuación balanceada si multiplicamos HNO₃ por 2.
N₂O₅ + H₂O ⇒ 2 HNO₃
b. En la segunda reacción, óxido de sodio reacciona con agua para formar hidróxido de sodio. Es una reacción de síntesis o combinación.
Na₂O + H₂O ⇒ NaOH
Podremos obtener la ecuación balanceada si multiplicamos NaOH por 2.
Na₂O + H₂O ⇒ 2 NaOH
Answer:
Number of moles = 10.6 mol
Explanation:
Given data:
Molar mass of H = 1.008 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Mass of citric acid = 2.03 kg (2.03×1000 = 2030 g)
Number of moles of citric acid = ?
Solution:
Formula:
Number of moles = mass/molar mass
Now we will calculate the molar mass of citric acid:
C₆H₈O₇ = (12.01× 6) + (1.008×8) + (16.00×7)
C₆H₈O₇ = 72.06 + 8.064+112
C₆H₈O₇ = 192.124g/mol
Number of moles = 2030 g/ 192.124g/mol
Number of moles = 10.6 mol
Answer:
Explanation:
Take a random sample of nuts from the jar. Let's take two handfuls, after shaking the jar and mixing the nuts thoroughly. Separate the nuts into almonds and cashews. Count each pile, then do the following calculation (these numbers are random, for example only).
<u> Count</u> <u>Percentage %</u>
Almonds 38 (38)/(87)x100
Cashews <u> 49</u> 49/87x100
87 87/87 = 100%
Ratio of Almonds to Cashews: <u>38/49</u>
Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:
2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
