Answer:
(C) through the atmosphere
Explanation:
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:
p and d orbitals
Explanation:
A π bond forms when two orbitals overlap side-on.
The most common types are formed by the overlap of p orbitals (Fig. 1).
However, d orbitals can also overlap sideways with each other to form dπ-dπ bonds and with p orbitals to form dπ-pπ bonds (Fig.2).
The half-life of the reaction is 50 minutes
Data;
- Time = 43 minutes
- Type of reaction = first order
- Amount of Completion = 45%
<h3>Reaction Constant</h3>
Let the initial concentration of the reaction be X
The reactant left = (1 - 0.45) X = 0.55 X = X
For a first order reaction
<h3>Half Life </h3>
The half-life of a reaction is said to be the time required for the initial amount of the reactant to reach half it's original size.
Substitute the values
The half-life of the reaction is 50 minutes
Learn more on half-life of a first order reaction here;
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