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DedPeter [7]
3 years ago
14

A student measures the length and mass of an object records the link is 45.2 cm convert distance measurement to the correct SI b

ase unit .4520.4524.52x10-4km17.8in
Chemistry
1 answer:
horrorfan [7]3 years ago
4 0

In science, physical quantities like length, weights etc are measured in units of metric or International System of units.

Some importance units are as follows:

1. Meter (m) for length.

2. Liter (L) for volume.

3. Gram (g) and kilogram (kg) for mass.

4. Degree Celsius (oC) for temperature.

5. Second for (s) time.

Given that the length of an object = 45.2 cm

SI unit of length is m

100 cm = 1.0 m

Therefore  

45.2 cm * 1 .0 m/100 cm

= 0.452 m  


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Ksivusya [100]
<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.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

PV=nRT

Where,

P is the pressure of the gas.

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:

Temperature(K)=Temperature(C\°) + 273K

So, we are given the following information:

Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K

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!

7 0
3 years ago
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