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jeka94
3 years ago
10

Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighter would have to be burned to produce 17

.9 L of carbon dioxide at STP
Chemistry
1 answer:
Hitman42 [59]3 years ago
5 0

<span>Answer is: mass of burned butane is 11.6 g.</span>

Chemical reaction: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.

m(butane) = 50,0 g.

<span> V(CO</span>₂) = 17,9 L.<span>
n(CO</span>₂) = V(CO₂) ÷ Vm.<span>
n(CO</span>₂) = 17,9 L ÷ 22,4 L/mol.<span>
n(CO</span>₂) = 0,8 mol.<span>
From chemical reaction n(CO</span>₂) : n(C₄H₁₀) = 8 : 2.<span>
n(C</span>₄H₁₀) = 0,8 mol ÷ 4.<span>
n(C</span>₄H₁₀) = 0,2 mol.<span>
m(C</span>₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).<span>
m(C</span>₄H₁₀) = 0,2 mol · 58 g/mol.<span>
m(C</span>₄H₁₀) = 11,6 g.

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Write a balanced chemical equation for the reaction that occurs when heptane, C7H16(l) , burns in air.
Pavlova-9 [17]
Start with Unbalanced Equation and balance it, so...
C7H16+O2--->CO2+H2O
There are 7 C atoms on the left-hand side, so we need 7 C atoms on the right-hand side. Add a 7 in front of the CO2...7CO2+H2O on right side now.
We have fixed 16 H atoms on the left-hand side, so we need 16 H atoms on the right-hand side. Add an 8 in front of H2O to make 16 (8x2)...7CO2+8H2O on right side now.
There are 22 O atoms on the right-hand side: 14 from the CO2 and 8 from the H2O. Add an 11 in front of the O2 on the left side to make 22 (11x2).
Every formula now has a fixed coefficient. You should have a balanced equation of...
C7H16+11O2--->7CO2+8H2O
8 0
3 years ago
A 2.2 M solution is made by with 0.45 moles of a solute. What is the final volume of this solution?
Savatey [412]

Answer: The final volume of this solution is 0.204 L.

Explanation:

Given: Molarity of solution = 2.2 M

Moles of solute = 0.45 mol

Molarity is the number of moles of solute present divided by volume in liters.

Molarity = \frac{no. of moles}{Volume (in L)}

Substitute the values into above formula as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\2.2 M = \frac{0.45}{Volume}\\Volume = 0.204 L

Thus, we can conclude that the final volume of this solution is 0.204 L.

7 0
3 years ago
Identify the oxidation state of Ba 2 + . +2 Identify the oxidation state of S in SO 2 . +2 Identify the oxidation state of S in
cupoosta [38]

Answer:

The oxidation state of Ba in cation Ba²⁺ is +2

The oxidation state of S in SO₂, is +4

The oxidation state of S in anion sulfate (SO₄⁻²) is +6

The oxidation state of Zn in the Zinc sulfate, is +2

Explanation:

We define oxidation state as the number which can be negative or positive that  

indicates the number of electrons that the atom has accepted or transferred.

All the elements in ground state has 0 as oxidation state.

This numbers are very important for redox reaction which are balanced by the ion electron method.

When the elements gain electrons, the element is being reduced so the oxidation state decreases.

When the elements release electrons, the element is oxidized so the oxidation state increases.

We have to think, that global charge of a compound is 0, for example in the ZnSO₄.

The sulfate anion has a global charge of -2 because it has released 2 protons, it came from the sulfuric acid (H₂SO₄). As the global charge is -2, oxygen acts with -2, and the anion has 4 atoms so the global charge of O is -8. Definetly S, has +6 as oxidation state.

In the SO₂, oxygen acts with -2 and there are 2 atoms in the compound, so the global charge is 0 and the global charge for O  is -4. Therefore S must act with +4.

Ba²⁺ is an element of group 2 and has a tendency to form a cation, so it can release electrons for that purpose.  At least, it can release 2 e⁻, that's why the oxidation state is +2. It can complete the octet rule and it will be isoelectronic with Xe.

3 0
3 years ago
Calculate the net force on the object below.
Anastaziya [24]

Answer:

first we add the same direction. 12N + 32 N=44N .

then we add the forces. 54 up + 44N down= 10N up

8 0
2 years ago
In a percentage composition investigation a compound was decomposed into its elements: 20.0 g of calcium, 6.0 g of carbon, and 2
inysia [295]

The percentage composition of this compound : 40%Ca, 12%C and 48%O

<h3>Further explanation</h3>

Given

20.0 g of calcium,

6.0 g of carbon,

and 24.0 g of oxygen.

Required

The percentage composition

Solution

Total mass of compound :

=mass calcium + mass carbon + mass oxygen

=20 g + 6 g + 24 g

=50 g

Percentage composition :

  • Ca-calcium

\tt \dfrac{20}{50}\times 100\%=40\%

  • C-carbon

\tt \dfrac{6}{50}\times 100\%=12\%

  • O-oxygen

\tt \dfrac{24}{50}\times 100\%=48\%

3 0
2 years ago
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