Isotopes of the same element
₃₅⁷⁷X and ₃₅⁸¹X
<h3>Further explanation</h3>
Given
Isotopes of element
Required
Isotopes of the same element
Solution
The elements in nature have several types of isotopes
Isotopes are elements that have the same Atomic Number (Proton) and different mass numbers
Element symbols that meet the 2 conditions above are :
₃₅⁷⁷X and ₃₅⁸¹X
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
The answer is letter C.
Because we classify something as a star when it is: a large ball of gas that undergoes nuclear fusion. Given this definition, a comet is not a star. A comet is a ball of ice and dirt hurtling through space, it shines only because it reflects ligh
Answer:
96.5 g/ml
Explanation:
If 5g is 19.3 then 25g is 19.3x5 which is 96.5 g/ml
Answer:
3.62x10⁻⁷ = Kb
Explanation:
The acid equilibrium of a weak acid, HX, is:
HX + H₂O ⇄ X⁻ + H₃O⁺
Where Ka = [X⁻] [H₃O⁺] / [HX]
And basic equilibrium of the conjugate base, is:
X⁻ + H₂O ⇄ OH⁻ + HX
Where Kb = [OH⁻] [HX] / [X⁻]
To convert Ka to Kb we must use water equilibrium:
2H₂O ⇄ H₃O⁺ + OH⁻
Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]
Thus, we can obtain:
Kw = Ka*Kb
Solving for Kb:
Kw / Ka = Kb
1x10⁻¹⁴ / 2.76x10⁻⁸ =
3.62x10⁻⁷ = Kb
