Answer : The mass of
needed are, 1.515 grams.
Explanation :
First we have to calculate the mole of
.

Now we have to calculate the moles of
.
The balanced chemical reaction will be,
produced from 1 mole of 
So, 0.005 mole of
produced from 0.005 mole of 
Now we have to calculate the mass of 


Therefore, the mass of
needed are, 1.515 grams.
A. water
the word hydro means water
Answer:
Carbon dioxide is added to the atmosphere by human activities. When hydrocarbon fuels (i.e. wood, coal, natural gas, gasoline, and oil) are burned, carbon dioxide is released. During combustion or burning, carbon from fossil fuels combine with oxygen in the air to form carbon dioxide and water vapor.Carbon moves from fossil fuels to the atmosphere when fuels are burned. When humans burn fossil fuels to power factories, power plants, cars and trucks, most of the carbon quickly enters the atmosphere as carbon dioxide gas. Each year, five and a half billion tons of carbon is released by burning fossil fuels.Carbon dioxide causes about 20 percent of Earth's greenhouse effect; water vapor accounts for about 50 percent; and clouds account for 25 percent.Likewise, when carbon dioxide concentrations rise, air temperatures go up, and more water vapor evaporates into the atmosphere—which then amplifies greenhouse heating
Leading up to this, calcium gave up 2 valence electrons and thus was denoted as a cation. These 2 electrons were transferred to bromine, which received an overall negative charge because of the addition of 2 valence electrons in its valence shell, and thus formed a negatively charged ion, an anion.
Both formed an ionic bond, due to the electrostatic charge of attraction between the 2 oppositely charged ions. If many ions of Ca and Br are present and numerous ionic bonds have formed it will undergo an arrangement which is that of an ionic lattice, type of structure.
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