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3 years ago

Write a balanced chemical equation for the reaction that occurs when heptane, C7H16(l) , burns in air.

1 answer:

8 0

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

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ivanzaharov [21]

Answer:

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Explanation:

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3 years ago

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cestrela7 [59]

Answer:

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3 years ago

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Which <br> molecule has the same shape and hybridization as methane

Korvikt [17]

Hi, the answer is <span>CF2Cl2 :)</span>

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3 years ago

A sealed container at 25oC contains a gas at a pressure of 104 kPa. What is the pressure of the gas when it is heated to 225oC?

seropon [69]

Answer:

174 kPa

Explanation:

Given that,

Initial temperature, T₁ = 25° C = 25+273 = 298 K

Final temperature, T₂ = 225°C = 225 + 273 = 498 K

Initial pressure, P₁ = 104 kPa

We need to find the new pressure. The relation between the temperature and pressure is given by :

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

So,

$P_2=\dfrac{P_1T_2}{T_1}\\\\P_2=\dfrac{104\times 498}{298}\\\\P_2=173.79\ kPa$

P₂ = 174 kPa

So, the new pressure is 174 kPa.

4 0

2 years ago

In the following reaction, which component acts as an oxidizing agent? 10 I− (aq) + 2 MnO4− (aq) + 16 H+ (aq) → 5 I2 (s) + 2 Mn2

antiseptic1488 [7]

Answer:

The oxidizing agent is the MnO₄⁻

Explanation:

This is the redox reaction:

10 I⁻ (aq) + 2 MnO₄⁻ (aq) + 16 H⁺ (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H2O (l)

Let's determine the oxidation and the reduction.

I⁻ acts with -1 in oxidation state and changes to 0, at I₂.

All elements in ground state has 0 as oxidation state.

As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.

In the permanganate (MnO₄⁻), Mn acts with +7 in oxidation state and decreased to Mn²⁺. As the oxidation state is lower, we talk about the reduction. Therefore, the permanganate is the oxidizing agent because it oxidizes iodide to iodine

3 0

3 years ago