Answer:
504.57 K.
Explanation:
From the question given above, the following data were obtained:
Number of mole (n) = 6.81 moles
Pressure (P) = 2.99 atm
Volume (V) = 94.35 L
Gas constant (R) = 0.0821 atm.L/Kmol
Temperature (T) =.?
Using the ideal gas equation, the temperature of the ideal gas can be obtained as follow:
PV = nRT
2.99 × 94.35 = 6.81 × 0.0821 × T
282.1065 = 0.559101 × T
Divide both side by 0.559101
T = 282.1065 / 0.559101
T = 504.57 K.
Thus, the temperature of the ideal gas is 504.57 K.
c) potassium, K(s)
b) manganese, Mn(s)
d) boron, B(s)
Answer:
The molar mass of:
Helium = 4.00 g/mol
Potassium = 39.0983 g/mol
Manganese = 54.94 g/mol.
Boron = 10.81 g / mol
Explanation:
Helium = 4.00 g/mol
Potassium = 39.0983 g/mol
Manganese = 54.94 g/mol.
Boron = 10.81 g / mol
Answer:
7.97 mol Ar
Explanation:
Use Avogadro's number to convert atoms to moles. This number is the number of atoms in one mole. There are 6.022 × 10²³ atoms in one mole.
Divide the number of atoms given by Avogadro's number.
(4.80 × 10²⁴)/(6.022 × 10²³) = 7.97 mol
There are 7.97 moles of argon.
Answer:
3.0 moles (A P E X)
Explanation:
Answer:
M
Explanation:
Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.
Partial pressure = Henry's constant x molar concentration
0.21 = Hx1.38x
H =
H = 152.17 atm/M
For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:
P = 0.21*0.875 = 0.1837 atm
Then, the molar concentration [O2], will be:
P = Hx[O2]
0.1837 = 152.17x[O2]
[O2] = 0.1837/15.17
[O2] = M
The molar concentration of O2 in the surface water of a mountain lake at 20 °C and an atmospheric pressure of 665 torr is approximately 1.21×10-3 M.
To calculate the molar concentration of O2 in the surface water of a mountain lake using Henry's law, we first need to understand how pressure affects the solubility of gases and vice versa.
As per Henry's law, at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The partial pressure of O2 in air at sea level is 0.21 atm. This means that when the atmospheric pressure is 1 atm, the molar concentration of O2 is 1.38×10−3 M. At higher altitudes, the atmospheric pressure reduces. The given atmospheric pressure at the mountain lake is 665 torr, which is approximately 0.875 atm.
Using these values in Henry's law, the molar concentration of O2 can be calculated as:
C = P * x
where C is molar concentration, P is atmospheric pressure, and x is given solubility at 1 atm. Substituting the values:
C = (0.875 atm) * (1.38×10−3 M) = 1.21x10-3 M approximately
#SPJ12
Answer:
Explanation:
I need more information to answer this question and a better figure.
1. What is the frequency of the NMR machine?
Possible solution:
1. coupling constant Jab (in ppm) is given by
\nu_{a} - \nu_{b} = 4J_{ab}
2. Jab (in ppm) * Frequency of machine in (MHz)/106 is Jab in Hz
3. for cis vicinal Hydrogen Jab = 6-14 Hz
4. for trans vicinal Hydrogens Jab = 11-18 Hz
Now, considering 2 doublets are centered at 7.14 and 7.28 ppm , it gives
7.14 -7.28 = 4 Jab
thus, Jab = 0.07 ppm
Now if we consider a 100 MHz machine,
Jab = 7 Hz , thus indicating cis product
but if machine is 300 MHz
then Jab = 21 Hz , thus indicating a trans product.
But, most probably I feel it is a trans product. I hope it helps.
The atoms having the same number of protons but a different number of neutrons have been termed isotopes.
The neutrons and protons are the constituents of the nucleus. The number of protons and electrons is equal in the atom. When two different atoms have the same number of protons the species have been termed isotopes. The atomic number of the species differ that resulting in the different positions in the periodic table.
For more information about the atoms, refer to the link:
Among all three PCl₃, NO₃⁻ , I₃⁻, H₂Se only I₃⁻ will involve participation of d-orbitals in hybridization as bonding in I₃⁻ will include, s, p and d orbital as shown in the image attached. there will be sp³d hybridization, there will be presence of three lone pairs and 2 bonds as
I⁻ has 8 valence electrons and 2 neighbours atoms which will need one electron each to satisfy their valency.
so the number of electrons on central atom will be:
8-1-1=6
That 6 electrons will make 6/2 =3 lone pairs.