Answer:
8740 joules are required to convert 20 grams of ice to liquid water.
Explanation:
The amount of heat required (
), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:
![Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}]](https://tex.z-dn.net/?f=Q%20%3D%20m%5Ccdot%20%5Bc%5Ccdot%20%28T_%7Bf%7D-T_%7Bo%7D%29%2BL_%7Bf%7D%5D)
(1)
Where:
- Mass, measured in grams.
- Specific heat of ice, measured in joules per gram-degree Celsius.
, 
- Temperature, measured in degrees Celsius.
- Latent heat of fussion, measured in joules per gram.
If we know that 
, 
, 
, 
and 
, then the amount of heat is:
![Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}](https://tex.z-dn.net/?f=Q%20%3D%20%2820%5C%2Cg%29%5Ccdot%20%5Cleft%5C%7B%5Cleft%282.06%5C%2C%5Cfrac%7BJ%7D%7Bg%5Ccdot%20%5E%7B%5Ccirc%7DC%7D%20%5Cright%29%5Ccdot%20%5B0%5C%2C%5E%7B%5Ccirc%7DC-%28-50%5C%2C%5E%7B%5Ccirc%7DC%29%5D%2B334%5C%2C%5Cfrac%7BJ%7D%7Bg%7D%20%5Cright%5C%7D)
8740 joules are required to convert 20 grams of ice to liquid water.
Answer:
160.32 grams of Ca or 160 if rounded
Explanation:
Multiply moles of Ca by the conversion factor (molar mass of calcium) 40.08 g Ca/ 1 mol Ca, which then allows the cancelation of moles, leaving grams of Ca.
4 mol*40.08g/mol = 160.32 grams of Ca
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Answer:
cfcl3
Explanation:
When the cfcl3 in the atmosphere is hit by UV rays from the sun, it decomposes to Cl atoms. These Cl atoms then react with the ozone producing O2 and ClO. The ClO bombard with O atoms splitting the ClO up and the free Cl atom again bonds to another O3 molecule hence reducing it. One ClO can reduce several O3 molecules (up to 100,000) creating an ozone hole.
2Cl + O3 → 2ClO + 2O2
2ClO + 2O → 2Cl + 2O2
Answer: option B. 0.020 mole
Explanation:Please see attachment for explanation
Answer:
Option 2, Half of the active sites are occupied by substrate
Explanation:
Michaelis-Menten expression for enzyme catalysed equation is as follows:
![V_0=\frac{V_{max\ [S]}}{k_M+[S]}](https://tex.z-dn.net/?f=V_0%3D%5Cfrac%7BV_%7Bmax%5C%20%5BS%5D%7D%7D%7Bk_M%2B%5BS%5D%7D)
Here, 
is Michaelis-Menten constant and [S] is substrate concentration.
When [S]=Km
Rearrange the above equation as follows:
![V_0=\frac{V_{max}[S]}{k_M+[S]}\\V_0=\frac{V_{max}[S]}{[S]+[S]}\\V_0=\frac{V_{max}[S]}{2[S\\]}\\V_0=\frac{V_{max}}{2}](https://tex.z-dn.net/?f=V_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7Bk_M%2B%5BS%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7B%5BS%5D%2B%5BS%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7B2%5BS%5C%5C%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%7D%7B2%7D)
when [S]=Km, the rate of enzyme catalysed reaction becomes half of the maximum rate, that means half of the active sites are occupied by substrate.
Therefore, the correct option is option 2.
