The process releases energy and consequently we classify it as exothermic. (d) a person running releases warmness as muscle mass paintings. therefore, the procedure is exothermic.
d) Exothermic, heat is released as a person runs and muscle groups perform work.
A chemical response or bodily exchange is exothermic if warmth is released by using the system into the environment. Because the surroundings are gaining heat from the system, the temperature of the surroundings increases. The sign of q for an exothermic procedure is -ve due to the fact the device is dropping heat.
Some other examples of exothermic reactions:
1)Snow Formation in Clouds.
2)Burning of a Candle.
3)Rusting of Iron.
4)Formation of Ion Pairs.
5)response of strong Acid and Water. etc...
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Answer:
Substances 1 and 2
Explanation:
an element only has 1 kind of atoms :3
Answer:
When atoms join together to form molecules, they are held together by chemical bonds. These bonds form as a result of the sharing or exchange of electrons between the atoms. It is only the electrons in the outermost shell that ever get involved in bonding.
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.
