Answer:
S(metal) = 0.66J/g°C
Explanation:
We can find specific heat of a material, S, using the equation:
q = m*S*ΔT
<em>Where q is change in heat, m is the mass of the substance, S specific heat and ΔT change in temperature.</em>
The heat given by the metal is equal to the heat that water absorbs, that is:
m(Metal)*S(metal)*ΔT(Metal) = m(Water)*S(water)*ΔT(water)
<em>Where:</em>
m(Metal) = 76.0g
S(metal) = ?
ΔT(Metal) = 96.0°C-31.0°C = 65.0°C 
m(Water) = 120.0g
S(water) = 4.184J/g°C
ΔT(water) = 31.0°C-24.5°C = 6.5°C
Replacing:
76.0g*S(metal)*65.0°C = 120.0g*4.184J/g°C*6.5°C
S(metal) = 0.66J/g°C
<em />
The law of conservation applies because the energy is not been created or destroyed. The energy that the metal gives is absorbed by the water.
 
        
             
        
        
        
Answer : The volume of hydrogen gas at STP is 4550 L.
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,

where,
 = initial pressure of gas = 100.0 atm
 = final pressure of gas at STP = 1 atm
 = initial volume of gas = 50.0 L
 = final volume of gas at STP = ?
 = initial temperature of gas = 
 = final temperature of gas at STP = 
Now put all the given values in the above equation, we get:


Therefore, the volume of hydrogen gas at STP is 4550 L.
 
        
             
        
        
        
The cost of one antacid is 2.325 cents per tablet.
<u>Explanation:</u>
As per the question based on the student analysis we know that,
Total antacid tablets in a bottle = 120
Purchase Price of a bottle = $ 2.79
Cost of 1 antacid tablet 
  
As we know $1 = 100 cent
The cost of 1 antacid tablet = 
 × 100 cents = 2.325 cents/tablet
. 
Thus we came to know that it costs 2.325 cents/tablet
. 
 
        
             
        
        
        
Answer:
After 2.0 minutes the concentration of N2O is 0.3325 M
Explanation:
Step 1: Data given
rate = k[N2O]
initial concentration of N2O of 0.50 M
k = 3.4 * 10^-3/s
Step 2: The balanced equation
2N2O(g) → 2 N2(g) + O2(g)   
Step 3: Calculate the concentration of N2O after 2.0 minutes
We use the rate law to derive a time dependent equation.
-d[N2O]/dt = k[N2O] 
ln[N2O] = -kt + ln[N2O]i
  ⇒ with k = 3.4 *10^-3 /s
 ⇒ with t = 2.0 minutes = 120s
 ⇒ with [N2O]i = initial conc of N2O = 0.50 M
ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)
ln[N2O] = -1.101
e^(ln[N2O]) = e^(-1.1011)
[N2O} = 0.3325 M
After 2.0 minutes the concentration of N2O is 0.3325 M