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Vlad [161]
4 years ago
14

1. How would the loss of tin oxide from the evaporating, due to spattering, etc. affect the empirical formula of your tin oxide?

Explain your answer.
Chemistry
1 answer:
Burka [1]4 years ago
6 0

Answer:

It will have no effect

Explanation:

The loss of tin oxide to evaporation will have no effect on the empirical formula of a compound.

The empirical formula of any compound is the simplest formula of that compound by which the combining atoms can be represented.

This formula is not affect by physical changes.

According to the law of constant composition  "all pure samples of the same compound have the same element in the same proportion by mass".

Regardless of the mass loss or gain of any tin oxide compound, it will have the same empirical and molecular formula. The atoms are still combining in the ratio to give the product.

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Describe what happens in the cell before the chemical reaction.
Zina [86]

___________________________________________________________

Answer:

Chemical reactions that take place inside living things are called biochemical reactions. The sum of all the biochemical reactions in an organism is referred to as metabolism. Metabolism includes both exothermic (heat-releasing) chemical reactions and endothermic (heat-absorbing) chemical reactions.

___________________________________________________________

6 0
3 years ago
Pentane is a small liquid hydrocarbon, between propane and gasoline in size at C5H12. The density of pentane is 0.626 g/mL and i
Goshia [24]

Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and 3.073\times 10^4kJ/L respectively.

Explanation :

Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).

As we are given that:

\Delta H^o_{comb}=-3535kJ/mol

Fuel value = \frac{\Delta H^o_{comb}}{\text{Molar mass of pentane}}

Molar mass of pentane = 72 g/mol

Fuel value = \frac{3535kJ/mol}{72g/mol}

Fuel value = 49.09 kJ/g

Now we have to calculate the fuel density of pentane.

Fuel density = Fuel value × Density

Fuel density = (49.09 kJ/g) × (0.626g/mL)

Fuel density = 30.73 kJ/mL = 3.073\times 10^4kJ/L

Thus, the fuel density of pentane is 3.073\times 10^4kJ/L

4 0
4 years ago
Problem 2 - The mass of Earth is 6.0x1024 kg. The mass of a hydrogen atom is
n200080 [17]

Answer: has diffused more quickly, so it must be a

lighter gas. Its particles have greater velocity

than the particles of Y at the same temperature.

(Note though that they will both have the same

value for average kinetic energy.)

7 From the kinetic molecular theory we would

expect a solid to be more dense than its liquid,

and therefore that ice would sink in water.

8 Bubbles will be present through the volume

of the liquid. A brown gas is visible above the

brown liquid. As the two states are at the same

temperature, the particles have the same average

kinetic energy and are moving at the same

speed. The inter-particle distances in the gas are

significantly larger than those in the liquid.

9 At certain conditions of low temperature and low

humidity, snow changes directly to water vapour

by sublimation, without going through the liquid

phase.

10 Steam will condense on the skin, releasing energy

as it forms liquid at the same temperature (E–D

on Figure 1.4). This is additional to the energy

released when both the boiling water and the

condensed steam cool on the surface of the

Explanation:

4 0
3 years ago
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
melamori03 [73]

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

8 0
3 years ago
A gas has a volume of 240 mL and a pressure of 0.99 atm. If the pressure is decreased to 0.951 atm, what is the new volume of th
irakobra [83]

V2 = 250 ml

Explanation:

Given:

P1 = 0.99 atm. V1 = 240 ml

P2 = 0.951 atm. V2 = ?

We can use Boyle's law to solve for V2

P1V1 = P2V2

V2 = (P1/P2)V1

= (0.99 atm/0.951 atm)(240 ml)

= 250. ml

6 0
3 years ago
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