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3 years ago

What are 5 steps of balancing a chemical equation plz help me

Chemistry

1 answer:

nlexa [21]3 years ago

6 0

Answer:

Explanation:

Identify each element found in the equation. The number of atoms of each type of atom must be the same on each side of the equation once it has been balanced.

What is the net charge on each side of the equation? The net charge must be the same on each side of the equation once it has been balanced.

If possible, start with an element found in one compound on each side of the equation. Change the coefficients (the numbers in front of the compound or molecule) so that the number of atoms of the element is the same on each side of the equation. Remember, to balance an equation, you change the coefficients, not the subscripts in the formulas.

Once you have balanced one element, do the same thing with another element. Proceed until all elements have been balanced. It's easiest to leave elements found in pure form for last.

Check your work to make certain the charge on both sides of the equation is also balanced.

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2 years ago

A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of

jek_recluse [69]

<u>Answer:</u> The volume when the pressure and temperature has changed is $1.6\times 10^2mL$

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

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$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

Let us assume:

$P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K$

Putting values in above equation, we get:

$\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL$

Hence, the volume when the pressure and temperature has changed is $1.6\times 10^2mL$

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3 years ago