40 g NaOH. You must use 40 g NaOH to prepare 10.0 L of a solution that has a pH of 13.
<em>Step 1</em>. Calculate the pOH of the solution
pOH = 14.00 – pH = 14.00 -13 = 1
<em>Step 2</em>. Calculate the concentration of NaOH
[NaOH] = [OH^(-)] = 10^(-pOH) mol/L = 10^(-1) mol/L = 0.1 mol/L
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 10.0 L solution × (0.1 mol NaOH/1 L solution) = 1 mol NaOH
<em>Step 4</em>. Calculate the mass of NaOH
Mass of NaOH = 1 mol NaOH × (40.00 g NaOH/1 mol NaOH) = 40 g NaOH
<span>the formation of a gas
</span>
Sorry I cant I just need some points
Answer:
Iron remains = 17.49 mg
Explanation:
Half life of iron -55 = 2.737 years (Source)
Where, k is rate constant
So,
The rate constant, k = 0.2533 year⁻¹
Time = 2.41 years
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
= 32.2 mg
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
is the initial concentration
So,
![[A_t]=32.2\times e^{-0.2533\times 2.41}\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D32.2%5Ctimes%20e%5E%7B-0.2533%5Ctimes%202.41%7D%5C%20mg)
![[A_t]=32.2\times e^{-0.610453}\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D32.2%5Ctimes%20e%5E%7B-0.610453%7D%5C%20mg)
![[A_t]=17.49\ mg](https://tex.z-dn.net/?f=%5BA_t%5D%3D17.49%5C%20mg)
<u>Iron remains = 17.49 mg</u>
Q=m°C<span>ΔT
=(500g) x (1 cal/g.</span>°C) x (48°C-21°C) = 13500 cal
13500 cal / 1000 = 13.5 kcal
<span>"What is the caloric value (kcal/g) of the french fries?"
13.5 kcal/ 2.5 g = 5.4 kcal/g</span>
