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Agata [3.3K]
3 years ago
5

Balance the following equations:

Chemistry
1 answer:
GuDViN [60]3 years ago
8 0

<em><u>ANSWERS</u></em><em><u> </u></em><em><u>:</u></em><em><u>-</u></em>

<em>1</em><em>)</em>

\red{2}Na + Br2 → \red{2}NaBr

<em>2</em><em>)</em>

\blue{2}C4H6 + \blue{11}O2 → \blue{8}CO2 + \blue{6}H2O

<em>3</em><em>)</em>

<em>H3PO4 + \green{3}LiOH → \green{3}H2O + Li3PO4</em>

<em>4</em><em>)</em>

<em>Ca(NO3)2 + K2SO4 → CaSO4 + \pink{2}KNO3</em>

<em>5</em><em>)</em>

<em>\orange{2}H2O2→\orange{2}H2O+O2</em>

<em>(</em><em>mark</em><em> </em><em>me</em><em> </em><em>brainliest</em><em> </em><em>please</em><em> </em><em>I</em><em> </em><em>need</em><em> </em><em>3</em><em> </em><em>more</em><em> </em><em>:</em><em>)</em><em>)</em>

<em>\large\sf\red{thank \: you}</em>

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2 years ago
Need help !!!!! ASAP
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<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of CO_{2}

n=0.120mol

<h2>Why?</h2>

We are asked to calculate the number of moles of the given gas, also, we  are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

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PV=nRT

Where,

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V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

R=62.363\frac{mmHg.L}{mol.K}

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

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Now, isolating the number of moles, and substituting the given information, we have:

PV=nRT

n=\frac{PV}{RT}

n=\frac{PV}{RT}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole

Hence, we have that there were produced 0.120 moles of CO_{2}

n=0.120mol

Have a nice day!

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