All categories

2 years ago

What is the molecular mass of a gas whose density is 1.25 grams per liter at STP?

Chemistry

1 answer:

satela [25.4K]2 years ago

6 0

STP = standard temperature and pressure

Use the formula:

d = MP/RT

where:

d = density = 1.25 g/L

M = molar mass (unknown)

P = pressure = 1 atm (standard pressure)

R = universal gas constant = 0.8206 L*atm/mol*K

T = temperature = 0C or 293.15 K (standard temperature)

1.25 g/L = M*(1 atm)/(0.08206 L*atm/mol*K)*(273.15 K)

M = 28.0 grams/mol

You might be interested in

Rank the following from highest to lowest priority according to the Cahn-Ingold-Prelog system: -CH2CH3, -CHCH2, -CCH, -CH3.

mario62 [17]

Answer:

The order will be:

CCH > CHCH₂ > CH₂CH₃> CH₃

Explanation:

According to Cahn-Ingold-Prelog system we rank the groups based on the atomic number of directly attached atom with the chiral carbon.

For example: between C and H, we rank Carbon first.

If the same atoms are attached for different groups then we prioritized based on the second element with highest atomic number.

For example:

Among CH₃ and C₂H₅, the priority will be given to C₂H₅.

If an atom is double or triple bonded to the directly attached atom then each pi bond is considered to be a new atom.

Hence CH=CH₂ means, that there are two carbons attached to CH carbon.

So the order based on above selection rules will be:

CCH > CHCH₂ > CH₂CH₃> CH₃

7 0

2 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$