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Sholpan [36]
3 years ago
7

Will maynez burns a 0.8-g peanut beneath 76 g of water, which increases in temperature from 22âc to 46 âc. (the specific heat ca

pacity of water is 1.0 cal/(gââc).)
Chemistry
1 answer:
bekas [8.4K]3 years ago
7 0
Based on the information provided, it appears that you will need to calculate the amount of heat absorbed by the water from the peanut that was burned. We are given the following information:

specific heat capacity, c = 1.0 cal/g°C
mass of water = 76 g
Ti = 22°C
Tf = 46°C
change in temperature, ΔT = 24°C

We can use the formula q = mcΔT to measure the amount of energy absorbed by the water to increase in tempature:

q = (76 g)(1.0 cal/g°C)(24°C)

q = 1824 cal

Therefore, the water absorbed 1824 calories from the peanut that was burned.
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A distillation column with a partial reboiler and a total condenser is being used to separate a mixture of benzene, toluene, and
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Answer:

11.8.4 Distillation Columns

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The conditions of operation of the equipment associated with the distillation column, particularly the reboiler and bottoms pump, are severe, so that failure is more probable.

The reduction of hazard in distillation columns by the limitation of inventory has been discussed above. A distillation column has a large input of heat at the reboiler and a large output at the condenser. If cooling at the condenser is lost, the column may suffer overpressure. It is necessary to protect against this by higher pressure design, relief valves, or HIPS. On the other hand, loss of steam at the reboiler can cause underpressure in the column. On columns operating at or near atmospheric pressure, full vacuum design, vacuum breakers, or inert gas injection is needed for protection. Deposition of flammable materials on packing surfaces has led to many fires on opening of distillation column for maintenance.

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The protection of distillation columns is one of the topics treated in detail in codes for pressure relief such as APIRP 521. Likewise, it is one of the principal applications of trip systems.

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Rank these elements according to FIRST ionization energy, from highest to lowest:
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What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
lyudmila [28]

Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

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