Answer:
2,760 grams NaCl
Explanation:
To find grams of NaCl, you need to (1) convert moles of Na to moles of NaCl (via mole-to-mole ratio from reaction) and (2) convert moles of NaCl to grams (via molar mass from periodic table). The final answer should have 3 significant figures based on the given measurement.
2 Na + Cl₂ --> 2 NaCl
Molar Mass (NaCl) = 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl) = 58.44 g/mol
47.2 moles Na 2 moles NaCl 58.44 grams
---------------------- x --------------------------- x ------------------------- =
2 moles Na 1 mole NaCl
= 2,758.368 grams NaCl
= 2,760 grams NaCl
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Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
- CH3Cl + H2O → CH3OH + HCl
at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
- K(T) = A e∧(-Ea/RT)
- Ln K = Ln(A) - [(Ea/R)(1/T)]
∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
Answer:
The vapor pressure in solution is 0,0051 atm
Explanation:
This is the formula for vapor pressure lowering, the colligative property.
P vapor = Pressure sv pure . Xsv
Where Xsv is data.
Xsv means Molar fraction (moles solvent/total n° moles)
Vapor pressure of water, pure is 17.5 mmHg
P vapor = 0,0313 atm . 0163
P vapor in solution = 0,0051 atm
Molar fraction does not have units
A solution will have less vapor pressure than that observed in the pure solvent.