I know the decimal is .56 i hope someone knows the fraction
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Answer:
The procedure with the simple calorimeter to compare the energy released by different fuels is described below in detail.
Explanation:
If the response delivers heat (rxn < 0), then heat is consumed by the calorimeter (calorimeter > 0), and its temperature improves. Conversely, if the response consumes heat (rxn > 0), suddenly heat is carried from the calorimeter to the arrangement (calorimeter < 0), and the temperature of the calorimeter drops.
<span>Only if they have the same mass.</span>
Answer : The energy for vacancy formation in silver is, 
Explanation :
Formula used :
or,
So,
![N_v=[\frac{N_A\times \rho}{M}]\times e^{(\frac{-E}{K\times T})}](https://tex.z-dn.net/?f=N_v%3D%5B%5Cfrac%7BN_A%5Ctimes%20%5Crho%7D%7BM%7D%5D%5Ctimes%20e%5E%7B%28%5Cfrac%7B-E%7D%7BK%5Ctimes%20T%7D%29%7D)
where,
= equilibrium number of vacancies = 
E = energy = ?
M = atomic weight = 107.9 g/mole
= Avogadro's number = 
= density = 
T = temperature = 
K = Boltzmann constant = 
Now put all the given values in the above formula, we get:
![3.6\times 10^{20}L^{-1}=[\frac{(6.022\times 10^{23}mol^{-1})\times 9500g/L}{107.9g/mol}]\times e^{[\frac{-E}{(1.38\times 10^{-23}J/K)\times 1073K}]}](https://tex.z-dn.net/?f=3.6%5Ctimes%2010%5E%7B20%7DL%5E%7B-1%7D%3D%5B%5Cfrac%7B%286.022%5Ctimes%2010%5E%7B23%7Dmol%5E%7B-1%7D%29%5Ctimes%209500g%2FL%7D%7B107.9g%2Fmol%7D%5D%5Ctimes%20e%5E%7B%5B%5Cfrac%7B-E%7D%7B%281.38%5Ctimes%2010%5E%7B-23%7DJ%2FK%29%5Ctimes%201073K%7D%5D%7D)
Therefore, the energy for vacancy formation in silver is,
Answer:
The sun would appear to move more slowly across Mercury's sky.
Explanation:
This is because, the time it takes to do one spin or revolution on Mercury is 176 days (which is its period), whereas, the time it takes to do one spin or revolution on the Earth is 1 day.
Since the angular speed ω = 2π/T where T = period
So on Mercury, T' = 176days = 176 days × 24 hr/day × 60 min/hr × 60 s/min = 15,206,400 s
So, ω' = 2π/T'
= 2π/15,206,400 s
= 4.132 × 10⁻⁷ rad/s
So on Earth, T" = 1 day = 1 day × 24 hr/day × 60 min/hr × 60 s/min = 86,400 s
So, ω" = 2π/T"
= 2π/86,400 s
= 7.272 × 10⁻⁵ rad/s
Since ω' = 4.132 × 10⁻⁷ rad/s << ω" = 7.272 × 10⁻⁵ rad/s, <u>the sun would appear to move more slowly across Mercury's sky.</u>
