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mote1985 [20]
3 years ago
7

3.Write the chemical equation for the reaction when methane burns in (2)

Chemistry
1 answer:
IceJOKER [234]3 years ago
5 0

Answer:

a)CH4 + O2 = CO2 + H2O

b)4CH4+5O=2CO+8H2O+2C

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Which of the following is the main evidence of life in the early universe?
alukav5142 [94]
<span>A)photosynthetic bacteria</span>
3 0
4 years ago
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Determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. group of answer choices
lilavasa [31]

Considering the ideal gas law and the definition of molar mass, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

<h3>Ideal gas law</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that occupies.
  • T is its temperature.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Molar mass of the sample of gas</h3>

In this case you know:

  • P= 0.980 arm
  • V= 1.20 L
  • T= 287 K
  • R= 0.082 \frac{atmL}{molK}
  • n= ?

Replacing in the ideal gas law:

0.980 atm× 1.20 L= n× 0.082\frac{atmL}{molK}× 287 K

Solving:

(0.980 atm× 1.20 L)÷ (0.082\frac{atmL}{molK}× 287 K)= n

<u><em>0.04997 moles= n</em></u>

On the other hand, you know that the<u><em> mass of the sample of gas</em></u> is <u><em>0.458 grams</em></u>. Replacing in the definition of molar mass:

molar mass=\frac{0.458 grams}{0.04997 moles}

Solving:

<u><em>molar mass= 9.17 </em></u>\frac{g}{mol}

Finally, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

Learn more about

molar mass:

brainly.com/question/5216907

brainly.com/question/11209783

brainly.com/question/7132033

brainly.com/question/17249726

ideal gas law:

brainly.com/question/4147359

#SPJ1

6 0
2 years ago
The standard enthalpy of formation (ΔHf°) of calcium carbonate is –1207 kJ/mol. Which ONE of the equations below has ΔH° = –1207
Art [367]

Answer:

A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)

Explanation:

Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.

The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:

Ca(s), C(s) and O₂(g)

That means, the equation that represents standard enthalpy of CaCO₃ is:

<h3>A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)</h3><h3 />

<em>Is the equation that has ΔH° = -1207kJ/mol</em>

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3 years ago
What is conjugate solution​
mars1129 [50]

Answer:

Explanation: A mixture of two partially miscible liquids

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3 years ago
Why must the spot applied to a TLC plate be above the level of the developing solvent? What problem will ensue if the level of t
algol13

Answer:

See explanation

Explanation:

If the spot in TLC is below the solvent front, it will be observed that the spot, instead of being separated by the solvent as expected, will just dissolve away in the solvent and zero actual separation of the mixture is achieved.

If the solute is dissolved away instead of being separated by the solvent, then the experiment fails because no actual separation of the mixture is achieved.

Hence, in TLC, the spot must be applied above the solvent front so that the capillary movement of the solvent through the plate can lead to the eventual separation of the components of the mixture since the various components of the mixture will travel at different speeds through the plate.

Also, if the solvent is above the spot, the solvent may evaporate selectively from the points above the spot while separation is ongoing.

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