Answer:
1.096g
Explanation:
You must know the atomic mass of Hydrogen, Fluorine, and Sodium before you can start:
Hydrogen: 1.008g/mol
Fluorine: 18.99g/mol
Sodium: 22.98g/mol
Next, find the composition percentage of NaF
22.98 + 18.99 = 41.97
Fluorine is 18.99/41.97 =45.25%
Sodium is 100-45.25 = 54.75%
Ultimately we want to know about HF so find how much F is in 2.3g: 2.3 * 0.4525 = 1.041g
Find comp. percentage of HF
18.99+1.008 = 19.998; H/total F/total
Hydrogen 5.041%
Fluorine 94.959%
Laws of conservation of say we have 1.041g of fluorine in our HF. We know 1.041 is 94.959% of the mass of HF so do some simple math to find the remaining: 1.041/0.94959 = 1.096g
The geosphere and the atmosphere
Answer:
184.113 g/mol
Explanation: The atomic mass of Mg is 24.3 amu. The atomic mass of bromine is 79.9. Therefore, the formula weight of MgBr2 equals 24.3 amu + (2 × 79.9 amu), or 184.1 amu. Because a substance's molar mass has the same numerical value as its formula weight, the molar mass of MgBr2 equals 184.1 g/mol.
Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
Answer:
Decreases
Explanation:
The temperature decreases when latitude increases.