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EleoNora [17]
3 years ago
6

What is the MOLAR heat of combustion of methane(CH₄) if 64.00g of methane are burned to heat 75.0 ml of water from 25.00°C to 95

.00°C?(Water heat capacity-4.184 J/g°C)
Chemistry
1 answer:
melamori03 [73]3 years ago
8 0

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

Q = c \times m \times \Delta T

where,

  • c: specific heat capacity
  • m: mass
  • ΔT: change in the temperature

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol

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a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temp
Snowcat [4.5K]

The heat of solution is -51.8 kJ/mol

<h3>What is the heat of solution?</h3>

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH =  11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity =  4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

Learn more about heat of solution:brainly.com/question/24243878

#SPJ1

4 0
2 years ago
8
Roman55 [17]

Answer:

305 litres of NO gas will be produced from 916 L of NO₂

Explanation:

Given the balanced equation of the chemical reaction as follows:

3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)

Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.

Molar volume of a gas at STP is 22.4 L

Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas

From the mole ratio of NO₂ to NO in the equation of reaction,

Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas

Volume of 13.631 moles of NO gas = 13.631 × 22.4

Volume of NO gas produced = 305.334L

Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L

5 0
2 years ago
What happens in this circuit if one of the light bulbs burns out?
bulgar [2K]
4.The other light bulb will stay on and glow brightly.
4 0
2 years ago
1. For each of the molecules below, determine the electron geometry, molecule geometry, and bond
Alexxx [7]

Answer:

CCl4- tetrahedral bond angle 109°

PF3 - trigonal pyramidal bond angles less than 109°

OF2- Bent with bond angle much less than 109°

I3 - linear with bond angles = 180°

A molecule with two double bonds and no lone pairs - linear molecule with bond angle =180°

Explanation:

Valence shell electron-pair repulsion theory (VSEPR theory) helps us to predict the molecular shape, including bond angles around a central atom, of a molecule by examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement which tends to minimize repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom are either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far apart from each other as possible.

Lone pairs and multiple bonds are known to cause more repulsion than single bonds and bond pairs. Hence the presence of lone pairs or multiple bonds tend to distort the molecular geometry geometry away from that predicted on the basis of VSEPR theory. For instance CCl4 is tetrahedral with no lone pair and four regions of electron density around the central atom. This is the expected geometry. However OF2 also has four regions of electron density but has a bent structure. The molecule has four regions of electron density but two of them are lone pairs causing more repulsion. Hence the observed bond angle is less than 109°.

8 0
3 years ago
Will ya'll help me, please
JulsSmile [24]

Answer:

Either A or D

But i think D

hope this helps =3

sorry if im wrong

7 0
3 years ago
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