This is an incomplete question, here is a complete question.
If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________.
The reaction
It is a second-order reaction with a rate constant of 0.80 M⁻¹s⁻¹ at 11 °C.
Answer : The concentration of after 9.0 seconds is, 0.00734 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 0.80 M⁻¹s⁻¹
t = time taken = 142 second
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.0440 M
Putting values in above equation, we get:
![0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)](https://tex.z-dn.net/?f=0.80M%5E%7B-1%7Ds%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7B142s%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.0440M%29%7D%5Cright%29)
![[A]=0.00734M](https://tex.z-dn.net/?f=%5BA%5D%3D0.00734M)
Hence, the concentration of after 9.0 seconds is, 0.00734 M
We will use the formula q = mCΔT.
where q = heat energy released.
m= mass.
C= specific heat of water.
ΔT= change in temperature.
Specific heat of water has not been given so we will just use the standard value of that which is 4.186J/g °C.
q = mCΔT.
q = 200 × 4.186 × (50 -25)
q = 200 × 4.186 × 25
q = 5000 × 4.186
q = 20, 930
The total amount of energy released is 20, 930 J
Answer:
Solution of isopropanol is 10.25 molal
Explanation:
615 g of isopropanol (C3H7OH) per liter
We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.
Density of water 1g/mL → Density = Mass of water / 1000 mL of water
Notice we converted the L to mL
Mass of water = 1000 g (which is the same to say 1kg)
Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol → 615 g . 1mol / 60g = 10.25 moles
Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m
When you say the solution is hypertonic, it means that the solution has a higher osmotic pressure. The formula for this is:
P = iMRT,
for strong electrolytes, i = number of ions.
for nonelectrolytes, i = 1
1. The P for sucrose solution which is a nonelectrolyte (assuming room temp):
P = (1)(1m)(8.314 J/mol-K)(298 K)
P = 2477.572 Pa
The P for NaCl solution, which is a strong electrolyte:
P = (2)(1 m)(8.314)(298 K)
P = 4955.144 Pa
<em>So, that means that NaCl is more hypertonic than the sucrose solution.</em>
2. For the second question, the P for the combination of 1 m glucose (nonelectrolyte) and 1 m sucrose is:
P = (1)(1 m)(8.314)(298 K) + (1)(1)(8.314)(298 K) = 4955.144 Pa
<em>In this case, the osmotic pressures are now equal. It is not hypertonic, but isotonic.</em>
Answer:
The specific heat capacity of the metal is 0.843J/g°C
Explanation:
Hello,
To determine the specific heat capacity of the metal, we have to work on the principle of heat loss by the metal is equals to heat gained by the water.
Heat gained by the metal = heat loss by water + calorimeter
Data,
Mass of metal (M1) = 512g
Mass of water (M2) = 325g
Initial temperature of the metal (T1) = 15°C
Initial temperature of water (T2) = 98°C
Final temperature of the mixture (T3) = 78°C
Specific heat capacity of metal (C1) = ?
Specific heat capacity of water (C2) = 4.184J/g°C
Heat loss = heat gain
M2C2(T2 - T3) = M1C1(T3 - T1)
325 × 4.184 × (98 - 78) = 512 × C1 × (78 - 15)
1359.8 × 20 = 512C1 × 63
27196 = 32256C1
C1 = 27196 / 32256
C1 = 0.843J/g°C
The specific heat capacity of the metal is 0.843J/g°C
