This boy said he was finna destroy me and winked what does that mean

A gas sample occupies 3.25 liters at 297.5K and 2.4 atm. Determine the temperature at which the gas will occupy 4.25 L at 1.50 a

lorasvet [3.4K]

For equal moles of gas, temperature can be calculated from ideal gas equation as follows:

P×V=n×R×T ...... (1)

Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.

2.4 atm ×3.25 L=n×R×297.5 K

Rearranging,

n\times R=0.0262 atm L/K

Similarly at final pressure and volume from equation (1),

1.5 atm ×4.25 L=n×R×T

Putting the value of n×R in above equation,

1.5 atm ×4.25 L=0.0262 (atm L/K)×T

Thus, T=243.32 K

7 0

3 years ago

What determines the reactivity of a metal?

fgiga [73]

The reactivity of a metal is determined by how tightly the metal holds onto the electrons in the outermost energy levels (valence electrons)

7 0

3 years ago

The lowest layer of Earths atmosphere is the__

alina1380 [7]

Answer:

Troposphere

Explanation:

The troposphere is the lowest layer in the atmosphere, and where all the weather occurs. After the troposphere, there's the stratosphere, mesosphere, thermosphere, and exosphere.

I hope this helps!

7 0

3 years ago

Read 2 more answers

A chemist uses hot hydrogen gas to convert chromium(III) oxide to

Darya [45]

Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, $Cr_{2}O_{3}$

Explanation:

The reaction equation for given reaction is as follows.

$Cr_{2}O_{3} + 3H_{2} \rightarrow 2Cr + 3H_{2}O$

Here, 1 mole of $Cr_{2}O_{3}$ reacts with 3 moles of $H_{2}$ .

As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide $(Cr_{2}O_{3})$ is 152 g/mol.

Number of moles is the mass of substance divided by its molar mass. So, moles of $Cr_{2}O_{3}$ is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{76 g}{152 g/mol}\\= 0.5 mol$

Now, moles of $H_{2}$ .given by 0.5 mol of $Cr_{2}O_{3}$ is calculated as follows.

$0.5 mol Cr_{2}O_{3} \times \frac{3 mol H_{2}}{1 mol Cr_{2}O_{3}}\\= 1.5 mol H_{2}$

As molar mass of $H_{2}$ is 2.016 g/mol. Therefore, mass of $H_{2}$ is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\1.5 mol = \frac{mass}{2.016 g/mol}\\mass = 3.024 g$

Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, $Cr_{2}O_{3}$ .

7 0

3 years ago

The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate b

____ [38]

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

$\Delta S_{vap}=\frac{\Delta H_{vap}}{T}$

We can solve for the temperature as follows:

$T=\frac{\Delta H_{vap}}{\Delta S_{vap}}$

Thus, with the proper units, we obtain:

$T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C$

Hence, answer is approximately 100 °C.

Best regards.

6 0

3 years ago