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2 years ago

A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?

1 answer:

4 0

Let's assume that the gas has ideal gas behavior. 
Then we can use ideal gas formula,
PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.

P = 60 cm Hg = 79993.4 Pa
V = 125 mL = 125 x 10⁻⁶ m³

n = ?

R = 8.314 J mol⁻¹ K⁻¹
T = 25 °C = 298 K

By substitution,
79993.4 Pa x 125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K
n = 4.0359 x 10⁻³ mol

Hence, moles of the gas = 4.0359 x 10⁻³ mol

Moles = mass / molar mass

Mass of the gas = 0.529 g

Molar mass of the gas = mass / number of moles
= 0.529 g / 4.0359 x 10⁻³ mol
 = 131.07 g mol⁻¹

Hence, the molar mass of the given gas is 131.07 g mol⁻¹

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Under identical conditions of temperature and pressure, which of the following gasses is the densest: Ne, CO2, or Cl2?

Tanzania [10]

Cl2=3.17g/L
Ne=.901g/L
CO2=1.96g/l
therefore Cl2 is the densest gas under the given conditions.

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2 years ago

In a hexagonal-close-pack (hcp) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes is 1:1:2. What i

Alex777 [14]

Explanation:

In a hexagonal-close-pack (HCP) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes = 1 : 1 : 2

let the :

Number of lattice point = 1x.

Number of octahedral points = 1x

Number of tetrahedral points = 2x

If anions occupy the HCP lattice points and cations occupy half of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B = $\frac{2x}{2}=x$

The formula of the compound will be = $A_{1x}B_{1x}=AB$

If anions occupy the HCP lattice points and cations occupy all of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B = 2x

The formula of the compound will be = $A_{1x}B_{2x}=AB_2$

6 0

2 years ago

Some instant cold packs contain ammonium nitrate and a separate pouch of water. When the pack is activated by squeezing to break

lubasha [3.4K]

Answer:

$T_f=-7.81^0C$

Explanation:

Hello,

a) In this case, since the heat associated with the dissolution of ammonium nitrate is positive, such reaction is endothermic as it absorbs heat.

b) Now, for computing the temperature once the dissolution is done, we apply (considering that it is a cooling process):

$T_f=T_0-\frac{\Delta H}{mCp}$

Nonetheless, we should first compute the moles of the mixture as:

$n_{mix}=135.0gH_2O*\frac{1molH_2O}{18gH_2O}+50.0gNH_4NO_3*\frac{1molNH_4NO_3}{80gNH_4NO_3}=8.125mol$

Thus, the total absorbed heat is:

$\Delta H=25.4kJ/mol*8.125mol=206.375kJ$

Now, the temperature is:

$T_f=25.0^0C-\frac{25.4kJ}{(135.0+50.0)g*4.184x10^{-3}kJ/g^0C} \\\\T_f=-7.81^0C$

Best regards.

3 0

2 years ago

The balanced equation shows how sodium chloride reacts with silver nitrate to form sodium nitrate and silver chloride. NaCl + Ag

Verdich [7]

NaCl First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound Sodium = 22.989769 Chlorine = 35.453 Silver = 107.8682 Nitrogen = 14.0067 Oxygen = 15.999 Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results For NaCl 22.989769 + 35.453 = 58.44277 For AgNO3 107.8682 + 14.0067 + 3 * 15.999 = 169.8719 Now calculate how many moles of each substance by dividing the total mass by the molar mass For NaCl 4.00 g / 58.44277 g/mol = 0.068443 mol For AgNO3 10.00 g / 169.8719 g/mol = 0.058868 Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl

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2 years ago

Urea contains two nitrogen atoms. what is the source of these two nitrogen atoms?

Oksi-84 [34.3K]

The source of these two nitrogen atoms are ammonia (NH₃) from nitrogen compounds (mostly metabolism of amino acids) through which excess nitrogen is eliminated from organisms. This process is called urea cycle, which extracted nitrogenous wastes.
The liver forms it by combining two ammonia molecules with a carbon dioxide molecule.

3 0

2 years ago