Answer:
Explanation:
Assuming that all caculations are at normal pressure and -1.72°C :
Where
is the number of moles of hydrogen
is the mass of hydrogen
is the density of hydrogen
Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen gas according to the following balanced equation.
Sn(s)+2HF(g)→SnF2(s)+H2(g)
Sn(s)+2HF(g)→
SnF
2
(s)+
H
2
(g)
How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?
Step 1: List the known quantities and plan the problem.
Known
given: 75.0 g Sn
molar mass of Sn = 118.69 g/mol
1 mol Sn = 2 mol HF (mole ratio)
Unknown
mol HF
Use the molar mass of Sn to convert the grams of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single two-step calculation.
g Sn → mol Sn → mol HF
Step 2: Solve.
75.0 g Sn×1 mol Sn118.69 g Sn×2 mol HF1 mol Sn=1.26 mol HF
75.0 g Sn×
1
mol Sn
118.69
g Sn
×
2
mol HF
1
mol Sn
=1.26 mol HF
Step 3: Think about your result.
The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three significant figures because the given mass has three significant figures.
I think the answer is true.
Answer:
6 x 10⁵ kg Hg
Explanation:
The mass of mercury in the entire lake is found by multiplying the concentration of the mercury by the volume of the lake.
The volume of the lake is calculated in cubic feet:
V = (SA)x(depth) = (100mi²)(5280ft/mi)² x (20ft) = 5.57568 x 10¹⁰ ft³
Cubic feet are then converted to mL (1cm³=1mL)
(5.57568 x 10¹⁰ ft³) x (12in/ft)³ x (2.54cm/in)³ = 1.578856752 x 10¹⁵ mL
The mass of mercury is then found:
m = CV = (0.4μg/mL)(1g/10⁶μg)(1kg/1000g) x (1.578856752 x 10¹⁵ mL) = 6 x 10⁵ kg Hg
The answer is 0.492 because there are 100 meters in a hectometer, and 49.2/100 is 0.492
