Answer:
34.81 ×10²³ atoms
Explanation:
Given data:
Mass of water = 52.0 g
Number of atoms of hydrogen = ?
Solution:
Number of moles of water:
Number of moles = mass/molar mass
Number of moles = 52.0 g/ 18 g/mol
Number of moles = 2.89 mol
1 mole of water contain 2 mole of hydrogen.
2.89 mole of water contain 2.89× 2= 5.78 moles of hydrogen.
Number of atoms of hydrogen:
1 mole = 6.022×10²³ atoms
5.78 mol × 6.022×10²³ atoms / 1mol
34.81 ×10²³ atoms
Something like the motherboard or RAM...it’s a physical part of the computer
Answer:
See Explanation
Explanation:
These problems are based upon the Ideal Gas Law, PV = nRT. Variables are gas phase variables and include ...
Pressure (P)
Volume (V)
Mass (moles (n))
R = Gas Constant = 0.08206L·atm/mol·K
Temperature (T)
Ideal Gas Law => PV = nRT
Note => when using the gas constant, R, convert all dimensional units to match those of L·atm/mol·K.
Problem 1 => Given ...
P = nRT/V
V = 250 ml = 0.250 L
n = 3.2 moles
R = 0.08206L·atm/mol·K
T = 45°C = (45 + 273)K = 318K
P = nRT/V = (3.2mol)(0.08206L·atm/mol·K)(318K)/(0.250L) = 334atm = 334atm(760mmHg/atm) = 253,853mmHg ≅2.5 x 10⁵mmHg (2 sig. figs.)
Problem 2 = Given ...
P = 5.05atm
V = nRT/P
n = 4.25 moles
R = 0.08206L·atm/mol·K
T = 27°C = (27 + 273)K = 300K
V = (4.25mol)(0.08206L·atm/mol·K)(300K)/(5.05atm) = 20.7 Liters O₂(g)
Problem 3 = Given ...
P = 302 KPa = (302KPa)(0.01atm/KPa) = 3.02 atm
V = 4.5 Liters
n = PV/RT
R = 0.08206L·atm/mol·K
T = 15°C = (15 + 273)K = 288K
n = (3.02atm)(4.5L)/(0.08206L·atm/mol·K)(288k) = 0.58 mole O₂ = 0.58 mole x 32 g/mole = 18.4 grams O₂
Problem 4 = Given ...
P = 810mmHg = 810mmHg(1atm/760mmHg) = 1.07atm
V = 2.0 Liters
n = 0.56 mole N₂(g)
R = 0.08206L·atm/mol·K
T = PV/nR = (1.07atm)(2.0L)/(0.56mol)(0.08206L·atm/mol·K) = 46.6K = (46.6 - 273)°C = -226°C
Problem 5 = Given ...
P = 3.2 atm
V = 55 Liters
n = PV/RT
R = 0.08206L·atm/mol·K
T = 17°C = (17 + 273)K = 290K
n = (3.2atm)(55L)/(0.08206L·atm/mol·K)(290K) = 7.4 moles N₂(g)
B. Theory
The hypothesis that stand the test of time (often tested and never rejected) is called theory. A theory is supported by a great dealcof evidence.
<u>Answer:</u> The total vapor pressure of the solution is 81.3 mmHg
<u>Explanation:</u>
To calculate the total pressure, we use the equation given by Dalton and Raoults, which is:
![p_T=(p_A\times \chi_A)+(p_B\times \chi_B)](https://tex.z-dn.net/?f=p_T%3D%28p_A%5Ctimes%20%5Cchi_A%29%2B%28p_B%5Ctimes%20%5Cchi_B%29)
where,
= total vapor pressure
We are given:
Mole fraction of benzene = 0.580
Mole fraction of toluene = (1 - 0.580) = 0.420 (Total mole fraction is always equation to 1)
Vapor pressure of benzene = 183 mmHg
Vapor pressure of toluene = 59.2 mmHg
Putting values in above equation, we get:
![p_T=(183\times 0.580)+(59.2\times 0.420)\\\\p_T=81.3mmHg](https://tex.z-dn.net/?f=p_T%3D%28183%5Ctimes%200.580%29%2B%2859.2%5Ctimes%200.420%29%5C%5C%5C%5Cp_T%3D81.3mmHg)
Hence, the total vapor pressure of the solution is 81.3 mmHg