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.
Answer:
a) distance is 4+7+1+8=20 blocks
b) displacement is 10 blocks
Explanation:
find displacement: x and y
x axis displacement = 4-1 = 3 blocks
y axis displacement = -7+8= 1 block
displacement = the square root of 3^2 + 1^2
= 9+1 = 10 blocks.
You can find the angle of displacement with respect to the initial position using trig identities, if you wish.
Answer:
Total partial pressure, Pt = 821 mm Hg
Partial pressure of Helium, P1 = 105 mm Hg
Partial pressure of Nitrogen, P2 = 312 mm Hg
Partial pressure of Oxygen, P3 = ? mm Hg
According to Dalton's law of Partial pressures,
Pt = P1 + P2 + P3
So, <u>P3 = 404 mm Hg</u>
Answer:
See the attached image
Explanation:
The first step is the production of the <u>carboanion</u> in the 
compound. We will get the <u>negative charge</u> on the methyl group and the <u>positive charge</u> in the Li atom.
Then the carboanion can <u>attack the acetone</u>. The double bond of the oxo group would <u>delocalized</u> upon the oxygen, generating a positive charge in the carbon that can be attacked by the carboanion formaiting a <u>new C-C bond</u>.
