17. ΔH rxn is the enthalpy of a reaction. It is the amount of energy or heat absorbed in a reaction. If enthalpy is positive, it means the reaction absorbs heat, which means it is endothermic. If the enthalpy is negative, it means the reaction release heat, which means it is exothermic.
18. yes, it is possible in theory but it is not necessary. Water is the ideal, cheaper, and most abundant liquid for a calorimeter.
19. Specific heat= heat/mass*Temp. the mass is already known You can place the piece of metal in a calorimeter filled with water. the piece of metal and water must be at different temperatures. Ideally, you would heat up the water and let it cool down. This change in temperature in the temperature that goes into the formula for the piece of metal. The only missing value is the heat which can be easily calculated because water' specific heat is known which can be used to calculate the heat loss by the water, which is the same as the heat gain by the piece of metal. With all the three values calculated and measured, you can simply plug them into the formula and solve for the specific heat of the metal.
1. 100N to the right
2. 10N to the left
3. 0N they are balanced
4. 0N they are balanced
Just remember how many electrons can each sublevel hold.
S=2
P=6
D=10
Since we have 10 for atomic number, we can assume we have 10 electrons
1S2
2S2
2P6
The rest have zero because we already have 10 (2+2+6=10)
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.
