<u>Answer:</u> The mass difference between the two is 7.38 grams.
<u>Explanation:</u>
To calculate the number of moles, we use the equation given by ideal gas follows:
where,
P = pressure = 125 psi = 8.50 atm (Conversion factor: 1 atm = 14.7 psi)
V = Volume = 855 mL = 0.855 L (Conversion factor: 1 L = 1000 mL)
T = Temperature = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
R = Gas constant = 
n = number of moles = ?
Putting values in above equation, we get:
To calculate the number of moles, we use the equation:
.....(1)
Moles of air = 0.297 moles
Average molar mass of air = 28.8 g/mol
Putting values in equation 1, we get:
Mass of air, 
= 8.56 g
Moles of helium = 0.297 moles
Molar mass of helium = 4 g/mol
Putting values in equation 1, we get:
Mass of helium, 
= 1.18 g
Calculating the mass difference between the two:
Hence, the mass difference between the two is 7.38 grams.
Answer:
The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is 12.90% by mass
Explanation:
2.23 M aqueous solution of NaCl means there are 2.23 moles of NaCl in 1000 mL of solution.
We know that density is equal to ratio of mass to volume.
Here density of solution is 1.01 g/mL.
So mass of 1000 mL solution = (
) g = 1010 g
molar mass of NaCl = 58.44 g/mol
So mass of 2.23 moles of NaCl = (
) g = 130.3 g
% by mass is ratio of mass of solute to mass of solution and then multiplied by 100.
Here solute is NaCl.
So % by mass of 2.23 M aqueous solution of NaCl = 
% = 12.90%
The Structure of Glycine is attached below and each central atom is encircled with different colors.
Molecular Shape around Nitrogen Atom (Orange):
As shown, Nitrogen is making three single bonds with two hydrogen atoms and one carbon atom hence, it has three bonded pair electrons and a single lone pair of electron. Therefore, according to VSEPR theory it has a tetrahedral electronic geometry but due to repulsion created by lone pair of electrons its molecular geometry becomes Trigonal Pyramidal.
Molecular Shape around Carbon Atom (Green):
As shown, Carbon is making four single bonds with two hydrogen atoms and one nitrogen atom one with carbon atom of carbonyl group hence, it has four bonded pair electrons. Therefore, according to VSEPR theory it has Tetrahedral geometry.
Molecular Shape around Carbon Atom (Blue):
As shown, Carbon is making two single bonds with oxygen and carbon atoms and a double bond with oxygen. Hence, it has a Trigonal Planar geometry.
Molecular Shape around Oxygen Atom (Red):
As shown, Oxygen is making two single bonds with one carbon atoms and one hydrogen atom hence, it has two bonded pair electrons and two lone pair of electrons. Therefore, according to VSEPR theory it has a tetrahedral electronic geometry but due to repulsion created by lone pair of electrons its molecular geometry becomes Bent.
Answer:
1 and 3.
Explanation:
The entropy measures the randomness of the system, as higher is it, as higher is the entropy. The randomness is associated with the movement and the arrangement of the molecules. Thus, if the molecules are moving faster and are more disorganized, the randomness is greater.
So, the entropy (S) of the phases increases by:
S solid < S liquid < S gases.
1. The substance is going from solid to gas, thus the entropy is increasing.
2. The substance is going from a disorganized way (the molecules of I are disorganized) to an organized way (the molecules join together to form I2), thus the entropy is decreasing.
3. The molecules go from an organized way (the atom are joined together) to a disorganized way, thus the entropy increases.
4. The ions are disorganized and react to form a more organized molecule, thus the entropy decreases.
Answer:
B. –99 kJ.
Explanation:
We have the following information:
1. C(s) + O₂(g) → CO₂(g);
ΔH = -393 kJ
2. 2CO(g) + O₂ → 2CO₂(g);
ΔH = -588 kJ
Using Hess's Law, Our target equation has C(s) on the left hand side, so we re-write equation 1:
1. C(s) + O₂(g) → CO₂(g);
ΔH = -393 kJ
So, we reverse equation 2 and divide by 2, we have equation 3:
3. CO₂(g) → CO(g) + ½O₂;
ΔH = +294 kJ
That is, change the sign of ΔH and divide by 2. Then we add equations 1 and 3 and their ΔH values.
This gives:
C(s) +½O₂(g) → CO(g);
ΔH = +294 - 393 kJ
= -99 kJ
The standard enthalpy of formation of carbon monoxide is -99 kJ/mol.
