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Hope this helps!
Answer:
Option B. 176g/mol
Explanation:
We'll begin by writing the chemical formula for hexasodium difluoride. This is given below:
Hexasodium means 6 sodium atom
Difluoride means 2 fluorine atom.
Therefore, the formula for hexasodium difluoride is Na6F2.
The relative formula mass of a compound is obtained by simply adding the atomic masses of the elements present in the compound.
Thus, the relative formula mass of hexasodium difluoride, Na6F2 can be obtained as follow:
Molar mass of Na = 23g/mol
Molar mass of F = 19g/mol
Relative formula mass Na6F2 = (23x6) + (19x2)
= 138 + 38
= 176g/mol
Therefore, the relative formula mass of hexasodium difluoride, Na6F2 is 176g/mol
Answer: this is not my own words :)
Learn how factors, such as exposure to the atmosphere, the composition of rock and climate affect the rate of weathering. Weathering is the mechanical or chemical process by which rocks are broken down. The weathering that happens to rocks can be understood by relating it to the digestion of a chocolate chip cookie
One of the factors that affects the rate of weathering is exposure to the atmosphere. ... Another factor that affects the rate of weathering is the composition of rock. Rocks are composed of minerals, which are solid substances found within rocks that have their own distinct chemical composition.
Answer:
5)0.53liters
6)1.634.6°c
Explanation:
5) v1=25liters, t1=12000°c. v2=?, t2=250°c
v2=(250×25)/12000
v2=6250/12000
v2=0.52liters
6)v1=130,v2=85, t1=2500, t2=?
t2=(85×2500)/130
t2=212500/130
t2= 1634.6°c
Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
<em />
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>
