PLEASE HELP ME !!!!!!!!!!!!!!

A 1.20-L container contains 1.10 g of an unknown gas at STP. What is the molecular weight of the unknown gas?

SOVA2 [1]

Answer:

M = 20.5 g/mol

Explanation:

Given data:

Volume of gas = 1.20 L

Mass of gas = 1.10 g

Temperature and pressure = standard

Solution:

First of all we will calculate the density.

Formula:

d = mass/ volume

d = 1.10 g/ 1.20 L

d = 0.92 g/L

Now we will calculate the molar mass.

d = PM/RT

0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K

M = 0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K / 1 atm

M = 20.5 g/mol

8 0

2 years ago

The state of matter with the slowest moving atoms is _

devlian [24]

The state of matter with the slowest moving atoms is solid!

6 0

2 years ago

After swimming in the ocean for several hours, the swimmers noticed that their fingers appeared to be very wrinkled or shriveled

MatroZZZ [7]

The answer is hypertonic. In osmosis, water molecules move from a hypotonic solution to the hypertonic solution, through a semipermeable membrane. This occurs until both solutions become isotonic relative to each other. In osmosis, only the movement of water molecules occurs since the ions are large enough to pass through the pores of the semipermeable membrane, in this case, the cell membrane. Due to loss of water in the process of osmosis, the cells in the fingers of the swimmers shrunk hence looked shriveled.

5 0

2 years ago

A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol

Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the solution

$c_1$ = specific heat of calorimeter = $12.1J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water or solution = $Density\times Volume=1/mL\times 100.0mL=100.0g$

$\Delta T$ = change in temperature = $T_2-T_1=(26.3-20.2)^oC=6.1^oC$

Now put all the given values in the above formula, we get:

$q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]$

$q=2623.61J$

Now we have to calculate the enthalpy change for the process.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = $Concentration\times Volume=1M\times 0.050L=0.050mole$

$\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole$

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

8 0

2 years ago

How would you properly measure the mass of a liquid

MArishka [77]

You can consider the density of the water. Thus, in order to properly measure the mass of a liquid, we can first get the volume and density of the liquid material or substance. We can firstly utilize the formula to get the mass from deriving the set formula of density to mass: Since density is mass over the volume, 
1. D=m/v 
2. We can transmute the formula to m = dv
3. Mass is density times the volume

4 0

2 years ago