Conjugate base pairs are acid and bases having common features. These features are the equal gain or loss of protons of the pairs. Conjugate pairs should always be one base and one acid. One would not exist without the other. Conjugate acids are the substances that gains protons while conjugates bases are those that loses protons. <span>The substances in the equilibrium reaction that is given is identified as follows:
HCO3^- + H2O <-----> CO3^2- + H3O^+
acid base conjugate base conjugate acid
HCO3^- ion is an intermediate molecule of CO2 and CO3^2-. When we add OH- to HCO3^-, we produce CO3^2-. And when we add H+ to HCO3, we produce CO2. </span>
Answer:
b) 2.0 mol
Explanation:
Given data:
Number of moles of Ca needed = ?
Number of moles of water present = 4.0 mol
Solution:
Chemical equation:
Ca + 2H₂O → Ca(OH)₂ + H₂
now we will compare the moles of Ca and H₂O .
H₂O : Ca
2 : 1
4.0 : 1/2×4.0 = 2.0 mol
Thus, 2 moles of Ca are needed.
Answer:
(a) 7.11 x 10⁻³⁷ m
(b) 1.11 x 10⁻³⁵ m
Explanation:
(a) The de Broglie wavelength is given by the expression:
λ = h/p = h/mv
where h is plancks constant, p is momentum which is equal to mass times velocity.
We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.
v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s
m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg
λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m
(b) For this part we have to use the uncertainty principle associated with wave-matter:
ΔpΔx > = h/4π
mΔvΔx > = h/4π
Δx = h/ (4π m Δv )
Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.
Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )
= 0.045 m/s
Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )
= 1.11 x 10⁻³⁵ m
This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.
Here is a picture of which shows you how many valence electrons are in the Lewis structure of xeo4
