Answer: hello the complete question is attached below
Visibility of molecular ion = m/z value of 77
Explanation:
For The molecular ion to be visible, it has to be at an m/z value of 77 and this is because molecular ions will have an m/z ratio = molecular mass of given molecule in most cases but not always in all cases.
And the visibility is possible after the removal of CH₃ ion.
ii) Evidence in the mass spectrum that suggests peak at m/z = 77
attached below
Answer: B!
Explanation: In the direction opposite of the direction of the applied force.There is no force in the other side being applied.
Answer:
E = 3.035× 10-¹⁹J = 1.9eV
f = 4.58 × 10¹⁴Hz
Explanation:
wavelength = 6.55 × 10-⁷m
c = 3 × 10⁸m/s
f = ?
E = ?
a) f = c/wavelength
f = 3 × 10⁸/6.55 × 10-⁷
f = 4.58 × 10¹⁴Hz
b) E = hc/wavelength
E = 6.626×10-³⁴ × 3 × 10⁸/ 6.55 × 10-⁷
E = 3.035 × 10-¹⁹J
1ev = 1.6 × 10-¹⁹J
Therefore E = 3.035/1.6 = 1.9eV
Answer:
Three half lives.
Explanation:
Given data:
Number of half lives passed = ?
Amount of parent isotope left = 12.5%
Solution:
At time zero = 100%
At 1st half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
At third half life = 25%/2 = 12.5%
It means three half lives would passed.
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
