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fgiga [73]
3 years ago
5

Methane is a major component of natural gas, which is used as a fuel in homes.Write a balanced equation for the complete oxidati

on reaction that occurs when methane (CH4) burns in air.
Chemistry
1 answer:
nexus9112 [7]3 years ago
5 0

Answer:

CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)

Explanation:

Methane is an hydrocarbon. Whenever hydrocarbon undergoes a reactio with oxygen, the products formed are carbon(iv) oxide and water.

The equation of the reaction is given as;

CH4 + O2 --> CO2 + H2O

Upon balancing the equation, we have;

CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)

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2
nignag [31]

Answer:

D

Explanation:

I think it is D. Think about it- if a human jumps, they are less than the gravitational force. But, if you are greater than the gravitational force, I think you will go into space.

4 0
3 years ago
NaOH(aq) is transferred into a test tube. CaCl2(aq) is added to the tube. What is the formula for the precipitate (if any) that
Ilya [14]

Answer:

2NaOH (aq) + CaCl2 (aq) -> 2NaCl(aq) + Ca(OH)2(s)

Formula of precipitate: Ca(OH)2 <em>(s)</em>

Explanation:

First, we do the double replacement reaction to determine our chemical equation between the reactants and products. Once we have our products, with a solubility chart (I added one below) we can determine which of the products is soluble or insoluble.

In this case NaCl is soluble or aqueous (meaning it can dissolve in water) and Ca(OH)2 is insoluble (meaning that when the reactions takes place, these two will form a solid/precipitate)

6 0
2 years ago
How many grams are in 4.5 moles of NaF?
Ilia_Sergeevich [38]

Answer:

3139542g

Explanation:

That's because if you talk about

4.5

moles of sodium fluoride, you would get a really absurd number of grams.

6 0
2 years ago
How does ionization energy change as you
kykrilka [37]

Answer:

How does the energy required to remove an electron from an atom change as you move left to right in Period 4 from potassium through iron? ... A greater nuclear charge pulls the electrons closer to the nucleus, decreasing the atomic radius.

3 0
2 years ago
A space air is at a temperature of 75 oF, and the relative humidity (RH) is 45%. Using calculations, find: (a) the partial press
earnstyle [38]

Answer:

A) Partial Pressure of dry air = 13.32 KPa

Partial Pressure of water vapour = 1.332 KPa

B) Humidity ratio; X = 0.0691

C) V_p = 0.8384 m³/Kg

Explanation:

A) We are given;

Temperature = 75°F

Relative Humidity = 45%

Now,to calculate the partial pressure, we will use the relationship;

Relative Humidity = (Partial Pressure/Vapour Pressure) × 100%

Making partial pressure the subject;

Partial Pressure = Relative Humidity × Vapour Pressure/100%

From the first table attached, at temperature of 75°F, the vapor pressure is 29.6 × 10^(-3) bar = 29.6 KPa

Thus;

Partial Pressure of dry air = (45 × 29.6)/100

Partial Pressure of dry air = 13.32 KPa

From online values, vapour pressure of water vapour at 75°F = 2.96 KPa

Thus;

Partial Pressure of water vapour = (45 × 2.96)/100 = 1.332 KPa

B) humidity ratio of moist air is given as;

X = 0.62198 pw / (pa - pw)  

where;

pw = partial pressure of the water vapor in moist air

pa = atmospheric pressure of the moist air

Thus;

X = (0.62198 × 1.332)/(13.32 - 1.332)  

X = 0.0691

C) Formula for moist air specific volume is;

V_p = (1 + (xRw/Ra) × RaT/p

Where;

V_p is specific volume

T is temperature = 75°F = 297.039 K

p is barometric pressure which in this case is standard sea level pressure = 101.325 KPa

pw is partial pressure of the water vapor in moist air = 1.332 KPa

Rw is individual gas constant for water = 0.4614 KJ/Kg.K

Ra is individual gas constant for air = 0.2869 KJ/Kg.K

V_p = (1 + (0.0691 * 0.4614/0.2869)) × 0.286.9 * 297.039/101.325

V_p = 0.8384 m³/Kg

6 0
3 years ago
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