Answer:
contain hereditary information
break down food into energy
Explanation:
Hereditary information is contained in genes and genes are found inside the cell. This implies that the cell contains hereditary information of organisms. This hereditary information is passed on during cell division from parent to daughter cells.
Metabolism occurs in the cells. The cells use oxygen to break down food materials to produce energy.
Answer:
The new partial pressures after equilibrium is reestablished:
![PCl_3,p_1'=6.798 Torr](https://tex.z-dn.net/?f=PCl_3%2Cp_1%27%3D6.798%20Torr)
![Cl_2,p_2'=26.398 Torr](https://tex.z-dn.net/?f=Cl_2%2Cp_2%27%3D26.398%20Torr)
![PCl_5,p_3'=223.402 Torr](https://tex.z-dn.net/?f=PCl_5%2Cp_3%27%3D223.402%20Torr)
Explanation:
![PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)](https://tex.z-dn.net/?f=PCl_3%28g%29%20%2B%20Cl_2%28g%29%5Crightleftharpoons%20PCl_5%28g%29%20)
At equilibrium before adding chlorine gas:
Partial pressure of the ![PCl_3=p_1=13.2 Torr](https://tex.z-dn.net/?f=PCl_3%3Dp_1%3D13.2%20Torr)
Partial pressure of the ![Cl_2=p_2=13.2 Torr](https://tex.z-dn.net/?f=Cl_2%3Dp_2%3D13.2%20Torr)
Partial pressure of the ![PCl_5=p_3=217.0 Torr](https://tex.z-dn.net/?f=PCl_5%3Dp_3%3D217.0%20Torr)
The expression of an equilibrium constant is given by :
![K_p=\frac{p_1}{p_1\times p_2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7Bp_1%7D%7Bp_1%5Ctimes%20p_2%7D)
![=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245](https://tex.z-dn.net/?f=%3D%5Cfrac%7B217.0%20Torr%7D%7B13.2%20Torr%5Ctimes%2013.2%20Torr%7D%3D1.245)
At equilibrium after adding chlorine gas:
Partial pressure of the ![PCl_3=p_1'=13.2 Torr](https://tex.z-dn.net/?f=PCl_3%3Dp_1%27%3D13.2%20Torr)
Partial pressure of the ![Cl_2=p_2'=?](https://tex.z-dn.net/?f=Cl_2%3Dp_2%27%3D%3F)
Partial pressure of the ![PCl_5=p_3'=217.0 Torr](https://tex.z-dn.net/?f=PCl_5%3Dp_3%27%3D217.0%20Torr)
Total pressure of the system = P = 263.0 Torr
![P=p_1'+p_2'+p_3'](https://tex.z-dn.net/?f=P%3Dp_1%27%2Bp_2%27%2Bp_3%27)
![263.0Torr=13.2 Torr+p_2'+217.0 Torr](https://tex.z-dn.net/?f=263.0Torr%3D13.2%20Torr%2Bp_2%27%2B217.0%20Torr)
![p_2'=32.8 Torr](https://tex.z-dn.net/?f=p_2%27%3D32.8%20Torr)
![PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)](https://tex.z-dn.net/?f=PCl_3%28g%29%20%2B%20Cl_2%28g%29%5Crightleftharpoons%20PCl_5%28g%29%20)
At initail
(13.2) Torr (32.8) Torr (13.2) Torr
At equilbriumm
(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr
![K_p=\frac{p_3'}{p_1'\times p_2'}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7Bp_3%27%7D%7Bp_1%27%5Ctimes%20p_2%27%7D)
![1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}](https://tex.z-dn.net/?f=1.245%3D%5Cfrac%7B%28217.0%2Bx%29%7D%7B%2813.2-x%29%2832.8-x%29%7D)
Solving for x;
x = 6.402 Torr
The new partial pressures after equilibrium is reestablished:
![p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr](https://tex.z-dn.net/?f=p_1%27%3D%2813.2-x%29%20Torr%3D%2813.2-6.402%29%20Torr%3D6.798%20Torr)
![p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr](https://tex.z-dn.net/?f=p_2%27%3D%2832.8-x%29%20Torr%3D%2832.8-6.402%29%20Torr%3D26.398%20Torr)
![p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr](https://tex.z-dn.net/?f=p_3%27%3D%28217.0%2Bx%29%20Torr%3D%28217%2B6.402%29%20Torr%3D223.402%20Torr)
The answer is 7. Valence electrons are the electrons in the very last shell, so we need to look at the outer “circle” and count the electrons, or the little black dots. There are 7 in the last shell.
Yes because look in the book dh
Answer:
3.91 minutes
Explanation:
Given that:
Biacetyl breakdown with a half life of 9.0 min after undergoing first-order reaction;
As we known that the half-life for first order is:
![t__{1/2}}= \frac{0.693}{k}](https://tex.z-dn.net/?f=t__%7B1%2F2%7D%7D%3D%20%5Cfrac%7B0.693%7D%7Bk%7D)
where;
k = constant
The formula can be re-written as:
![k = \frac{0.693}{t__{1/2}}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B0.693%7D%7Bt__%7B1%2F2%7D%7D)
![k = \frac{0.693}{9.0 min}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B0.693%7D%7B9.0%20min%7D)
![k = 0.077 min^{-1}](https://tex.z-dn.net/?f=k%20%3D%200.077%20min%5E%7B-1%7D)
Let the initial amount of butter flavor in the food be
= 100%
Also, the amount of butter flavor retained at 200°C
= 74%
The rate constant ![k = 0.077 min^{-1}](https://tex.z-dn.net/?f=k%20%3D%200.077%20min%5E%7B-1%7D)
To determine how long can the food be heated at this temperature and retain 74% of its buttery flavor; we use the formula:
![\frac{N_t}{N_0}= -kt](https://tex.z-dn.net/?f=%5Cfrac%7BN_t%7D%7BN_0%7D%3D%20-kt)
![t = - (\frac{1}{k}*In\frac{N_t}{N_0} )](https://tex.z-dn.net/?f=t%20%3D%20-%20%28%5Cfrac%7B1%7D%7Bk%7D%2AIn%5Cfrac%7BN_t%7D%7BN_0%7D%20%20%29)
Substituting our values; we have:
![t = - (\frac{1}{0.077}*In\frac{74}{100} )](https://tex.z-dn.net/?f=t%20%3D%20-%20%28%5Cfrac%7B1%7D%7B0.077%7D%2AIn%5Cfrac%7B74%7D%7B100%7D%20%20%29)
t = 3.91 minutes
∵ The time needed for the food to be heated at this temperature and retain 74% of its buttery flavor is 3.91 minutes