Answer:
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Explanation:
The balanced reaction is
Ca + 2H2O → Ca(OH)2 + H2
Here this reaction can be compared with
A + BC ---> AB + C
So here one reactant (A) is accepting a group which is being given by another compound (BC) however the A is not giving any group / element or ion
So this single displacement
Similarly in the given reaction
the anion OH- is only being replaced
The element Ca accepts OH- and H2O loses the same group to form new element H2
So the correct answer is
Single replacement also known as single displacement
Answer:
(a) r = 6.26 * 10⁻⁷cm
(b) r₂ = 6.05 * 10⁻⁷cm
Explanation:
Using the sedimentation coefficient formula;
s = M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle
s = M ( 1 - Vρ) / N*6πnr
making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs
Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s
r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)
r = 6.26 * 10⁻⁷cm
b. Using the formula r₂/r₁ = s₁/s₂
s₂ = 0.035 + 1s₁ = 1.035s₁
making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁
r₂ = 6.3 * 10⁻⁷cm / 1.035
r₂ = 6.05 * 10⁻⁷cm
Answer:
27.60 g urea
Explanation:
The <em>freezing-point depression</em> is expressed by the formula:
In this case,
- ΔT = 5.6 - (-0.9) = 6.5 °C
m is the molality of the urea solution in X (mol urea/kg of X)
First we<u> calculate the molality</u>:
- 6.5 °C = 7.78 °C kg·mol⁻¹ * m
Now we<u> calculate the moles of ure</u>a that were dissolved:
550 g X ⇒ 550 / 1000 = 0.550 kg X
- 0.84 m = mol Urea / 0.550 kg X
Finally we <u>calculate the mass of urea</u>, using its molecular weight:
- 0.46 mol * 60.06 g/mol = 27.60 g urea
