How many moles are in 454 Liters of a gas?

28. How many formula units are found in 0.250 moles of potassium nitrate?

Soloha48 [4]

Answer:

$\boxed {\boxed {\sf B. \ 1.5 *10^{23} \ formula \ units}}$

Explanation:

1 mole of any substance contains the same number of particles. The particles can vary (atoms, molecules, formula units), but there are always 6.022*10²³ particles. In this case, the particles are formula units of potassium nitrate or KNO₃.

Let's create a ratio.

$\frac {6.022*10^{23} \ formula \ units \ KNO_3}{1 \ mol \ KNO_3}$

Since we are trying to find the formula units in 0.250 moles, we multiply by that number.

$0.250 \ mol \ KNO_3 *\frac {6.022*10^{23} \ formula \ units \ KNO_3}{1 \ mol \ KNO_3}$

The units of moles of potassium nitrate cancel.

$0.250 *\frac {6.022*10^{23} \ formula \ units \ KNO_3}{1 }$

The denominator of 1 can be ignored, so we can make a simple multiplication problem.

$0.250 *{6.022*10^{23} \ formula \ units \ KNO_3}$

$1.5055 * 10^{23} \ formula \ units \ KNO_3$

If we round to the nearest tenth, the 0 in the hundredth place tells us to leave the 5 in the tenth place.

$1.5 *10^{23} \ formula \ units \ KNO_3$

0.250 moles of potassium nitrate is approximately equal to 1.5*10²³ formula units of potassium nitrate and choice B is correct.

7 0

3 years ago

Explain what is the difference<br> between the name of genus and the<br> name of species?

serious [3.7K]

The genus is a lower classification level that lies below family and above species . Species is the fundamental category of closely related organisms that lies below the genus .

4 0

4 years ago

Read 2 more answers

The final charge on each of the three separated spheres in part (b) is +3.0 μC. How many electrons would have to be added to one

marishachu [46]

Answer:

$1.873\times 10^{13}$ electrons should be added to one of these spheres to make it electrically neutral.

Explanation:

Total charge on each sphere = +3.0 μC = $3\times 10^{-6} C$

$1 \mu C=10^{-6} C$

In order to neutralize the positive charge equal magnitude of negative charge is to be added.

Total charge electrons, Q= -3.0 μC = $-3\times 10^{-6} C$

Number of electrons = n

Charge on a single electron, e = $-1.602\times 10^{-19} C$

Q = n × e

$-3\times 10^{-6} C=n\times (-1.602\times 10^{-19} C)$

$n=\frac{-3\times 10^{-6} C}{-1.602\times 10^{-19} C}=1.873\times 10^{13}$

$1.873\times 10^{13}$ electrons should be added to one of these spheres to make it electrically neutral.

4 0

3 years ago

Name the alkanes please help

Pie

Methane CH4 CH4 1 hexane C6H14 CH3(CH2)4CH3 5
ethane C2H6 CH3CH3 1 heptane C7H16 CH3(CH2)5CH3 9
propane C3H8 CH3CH2CH3 1 octane C8H18 CH3(CH2)6CH3 18
butane C4H10 CH3CH2CH2CH3 2 nonane C9H20 CH3(CH2)7CH3 35
pentane C5H12 CH3(CH2)3CH3 3 decane C10H22 CH3(CH2)8CH3 75

4 0

2 years ago

If measurements of a gas are 100L and 300 kilopascals and then the gas is measured a second time and found to be 75L, describe w

Tanya [424]

Answer:

The pressure of the gas increased (if temperature remained constant).

The Boyle's law supports this observation.

Explanation:

The initial measurements of the gas are given as;

volume = 100 L

Pressure = 300 kpa

The second measurement is given as;

Volume = 75 L

The second reading implies that the volume of the gas has decreased. If the temperature of the gas remained constant, then the pressure must have increased according to the Boyle's law;

At constant temperature, the pressure of a given mass of an ideal gas is inversely proportional to its volume.

4 0

3 years ago