Answer:
A = 674.33mmHg
B = 0.385atm
Explanation:
Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.
Mathematically,
P = kT, k = P / T
P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn
A)
Data:
P1 = 799mmHg
T1 = 50°C = (50 + 273.15) = 323.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2
P2 = (P1 × T2) / T1
P2 = (799 × 273.15) / 323.15
P2 = 674.37mmHg
The final pressure is 674.37mmHg
B)
P1 = 0.470atm
T1 = 60°C = (60 + 273.15)K = 333.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2,
P2 = (P1 × T2) / T1
P2 = (0.470 × 273.15) / 333.15
P2 = 0.385atm
The final pressure is 0.385atm
Answer:
c. 6.
Explanation:
Looking at the description given in the question, the elements involved must belong to the p- block of the periodic table and must be in period 5. They also must possess valence electrons in the 5p- orbital.
Now if we look at the p- block of period 5, the following elements satisfy these requirements; Sr, In, Sn, Sb, Te and I.
Hence there are six of such elements.
Answer:
2.12×10²³ atoms.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of zirconium also 6.02×10²³ atoms.
Thus, we can obtain the number of atoms present in 0.3521 mole of zirconium as follow:
1 mole of zirconium also 6.02×10²³ atoms.
Therefore, 0.3521 mole of zirconium will contain = 0.3521 × 6.02×10²³ = 2.12×10²³ atoms.
Therefore, 0.3521 mole of zirconium contains 2.12×10²³ atoms.
Answer:
1.123x10⁻⁴ moles of alanine
Explanation:
In order to convert grams of alanine into moles, <em>we need to know its molecular weight</em>:
The formula for alanine is C₃H₇NO₂, meaning <u>its molecular weight would be</u>:
- 12*3 + 7*1 + 14 + 16*2 = 89 g/mol
Then we <u>divide the sample mass by the molecular weight</u>, to do the conversion:
- 1.0x10⁻² g ÷ 89 g/mol = 1.123x10⁻⁴ moles
