Answer:
a) 381.2 g
b) 39916 g
c) 0.0013 lb mol
d) 29.6 g mol
Explanation:
The molecular weight (mw) of a compound is the mass of it per mole, so it's the ratio of the mass (m) per mole (n).
a) The molecular weight of one mol is found at the periodic table. So, for Mg, mw = 24.3 g/mol, for Cl = 35.5 g/mol, so for MgCl2, mw = 24.3 + 2*35.5 = 95.3 g/mol. The g mol is the mass divided by the molecular weight:
g mol = m/mw
4 = m/95.3
m = 381.2 g
b) The pound (lb) is a unity of mass, and the lb mol is a unity of the mass divided by the molecular weight. So, by the periodic table, the molecular weight of C3H8 is 3*12 (of C) + 8*1 (of H) = 44 lb/mol.
lb mol = m/mw
2 = m/44
m = 88 lb
1 lb = 453.592 g
So, m = 88*453.592 = 39916 g
c) The molecular weight of N2 is 2*14 (of N) = 28 lb/mol.
m = 16/453.592 = 0.0353 lb
lb mol = m/mw
lb mol = 0.0353/28
lb mol = 0.0013 lb mol
d) The molecular weight is 2*12 (of C) + 6*1(of H) + 1*16(of O) = 46 g/mol
3 lb = 1360.78 g
g mol = m/mw
g mol = 1360.78/46
g mol = 29.6 g mol
To find average atomic mass you multiply the mass of each isotope by its percentage, and then add the values up.
35 * 0.90 + 37 * 0.08 + 38 * 0.02 = 35.22
Average atomic mass closest to 35.22 amu.
Answer:
See explanation
Explanation:
Now , we have the equation of the reaction as;
2H2S(g) + 302(g)------->2SO2(g) + 2H2O(g)
This equation shows that SO2 gas is produced in the process. Let us recall that this same SO2 gas is the anhydride of H2SO4. This means that it can dissolve in water to form H2SO4
So, when SO2 dissolve in rain droplets, then H2SO4 is formed thereby lowering the pH of rain water. This is acid rain.
Answer:
A). The complementary shapes of an enzyme and a substrate.
Explanation:
The Lock-and-key mechanism was proposed by Emil Fischer for the first time and characterized as the metaphor which helps in elucidating the specificity of the enzymatic reactions. In this metaphor, the lock is described as the enzyme while 'key' is characterized as the substrate which the enzyme acts upon. If the key is not appropriately sized, it will not fit into the active site i.e. the keyhole of the lock or enzyme and reaction will not take place. Thus, <u>option A</u> is the correct answer.
Answer:
70.88 mL volume of 1.27 M of HCl is required.
Explanation:
Given data:
Initial volume = ?
Initial molarity = 1.27 M
Final volume = 197.4 mL
Final molarity = 0.456 M
Solution:
Formula:
M₁V₁ = M₂V₂
Now we will put the values in formula.
1.27 M × V₁ = 0.456 M × 197.4 mL
V₁ = 0.456 M × 197.4 mL/1.27 M
V₁ = 90.014M.mL/1.27 M
V₁ = 70.88 mL
70.88 mL volume of 1.27 M of HCl is required.
