Answer is: 5.22·10²² atoms of Iodine.
m(CaI₂) = 12.75 g; mass of calcium iodide.
M(CaI₂) = 293.9 g/mol; molar mass of calcium iodide.
n(CaI₂) = m(CaI₂) ÷ M(CaI₂).
n(CaI₂) = 12.75 g ÷ 293.9 g/mol.
n(CaI₂) = 0.043 mol; amount of calcium iodide.
In one molecule of calcium iodide, there are two iodine atoms
n(I) = 2 · n(CaI₂).
n(I) = 0.086 mol; amount of iodine atoms.
Na = 6.022·10²³ 1/mol; Avogadro number.
N(I) = n(I) · Na.
N(I) = 0.086 mol · 6.022·10²³ 1/mol.
N(I) = 5.22·10²²; number of iodine atoms.
Third quarter (or last quarter)
PH scale is from 1 to 14 and indicates how acidic or basic a solution is. To find pH or pOH we need to know the H⁺ ion concentration or OH⁻ concentration.
pH can be calculated using the following equation;
pH = -log[H⁺]
the H⁺ concentration of the given acid is 1.0 x 10⁻⁴ M. substituting this we can find the pH
pH = -log[1x10⁻⁴]
pH = 4
answer is 1) 4
1. The answer is option E, that is None of the above is correct.
As a polymer becomes more crystalline,
its melting point doesn't decreases, its density doesn't decreases, its stiffness doesn't decreases and its yield stress doesn't decreases.
2. The answer is option B, that is the molecules are arranged in sheets, with their long axes parallel and their ends aligned as well.
In the smectic A liquid-crystalline phase, molecules are arranged in sheets, with their long axes parallel and their ends aligned as well.
3. For a substitutional alloy to form, the two metals combined must have similar atomic radii and chemical bonding properties.
Answer:
C₅ H₁₂ O
Explanation:
44 g of CO₂ contains 12 g of C
30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .
18 g of H₂O contains 2 g of hydrogen
14.8 g of H₂0 will contain 1.644 g of H .
total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O
gram of O = 2.22
moles of C, O, H in the given compound = 8.236 / 12 , 2.22 / 16 , 1.644 / 1
= .6863 , .13875 , 1.644
ratio of their moles = 4.946 : 1 : 11.84
rounding off to digits
ratio = 5 : 1 : 12
empirical formula = C₅ H₁₂ O
