The theoretical yield of I2 in the reaction would be 0.23 g
<h3>Theoretical yield</h3>
This refers to the stoichiometric yield of a reaction.
From the equation of the reaction:
Ca(IO3)2 + 10 KI + 12 HCl → 6 I2 + CaCl2 + 10 KCl + 6 H2O
The mole ratio of Ca(IO3)2 and I2 is 1: 6
Mole of 15.00 mL, 0.0100 M Ca(IO3)2 = 15/1000 x 0.0100
= 0.00015 mole
Equivalent mole of I2 = 0.00015 x 6
= 0.009 mole
mass of 0.0009 I2 = 0.0009 x 253.809
= 0.23 g
More on stoichiometric calculations can be found here: brainly.com/question/6907332
Depends on what the base is. You would reference the base dissociation chart for that value.
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.
Answer:
A partial lunar eclipse occurs when the Earth moves between the Sun and Moon but the three celestial bodies do not form a straight line in space. When that happens, a small part of the Moon's surface is covered by the darkest, central part of the Earth's shadow, called the umbra.
Explanation:
Answer:
(i) specific heat
(ii) latent heat of vaporization
(iii) latent heat of fusion
Explanation:
i. Q = mcΔT; identify c.
Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.
ii. Q = mLvapor; identify Lvapor
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.
iii. Q = mLfusion; identify Lfusion
Here, Q is the heat, m is the mass and L is the latent heat of fusion.
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg solid into 1 kg liquid at constant temperature.
