Answer:
[CaSO₄] = 36.26×10⁻² mol/L
Explanation:
Molarity (M) → mol/L → moles of solute in 1L of solution
Let's convert the volume from mL to L
250 mL . 1L/1000 mL = 0.250L
We need to determine the moles of solute. (mass / molar mass)
12.34 g / 136.13 g/mol = 0.0906 mol
M → 0.0906 mol / 0.250L = 36.26×10⁻² mol/L
Answer:
It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.
Explanation:
A → B
Initial concentration of the reactant = x
Final concentration of reactant = 10% of x = 0.1 x
Time taken by the sample, t = ?
Formula used :
where,
= initial concentration of reactant
A = concentration of reactant left after the time, (t)
= half life of the first order conversion = 56.6 hour
= rate constant
Now put all the given values in this formula, we get
t = 188.06 hour
It will take 188.06 hours for the concentration of A to decrease 10.0% of its original concentration.
The answer is actually Water Splitting! Hope this works:D
Answer:
Water has a very high heat capacity, which makes it useful for radiators ... the temperature change that a given substance will undergo when it is either heated or cooled. ... The heat that is either absorbed or released is measured in joules. ... A 15.0 g piece of cadmium metal absorbs 134 J of heat while rising from 24.0°C to ...
To calculate the <span>δ h, we must balance first the reaction:
NO + 0.5O2 -----> NO2
Then we write all the reactions,
2O3 -----> 3O2 </span><span>δ h = -426 kj eq. (1)
O2 -----> 2O </span><span>δ h = 490 kj eq. (2)
NO + O3 -----> NO2 + O2 </span><span>δ h = -200 kj eq. (3)
We divide eq. (1) by 2, we get
</span>O3 -----> 1.5O2 δ h = -213 kj eq. (4)
Then, we subtract eq. (3) by eq. (4)
NO + O3 -----> NO2 + O2 δ h = -200 kj
- (O3 -----> 1.5 O2 δ h = -213 kj)
NO -----> NO2 - 0.5O2 δ h = 13 kj eq. (5)
eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)
O -----> 0.5O2 <span>δ h = -245 kj eq. (6)
</span>
Add eq. (6) to eq. (5), we get
NO -----> NO2 - 0.5O2 δ h = 13 kj
+ O -----> 0.5O2 δ h = -245 kj
NO + O ----> NO2 δ h = -232 kj
<em>ANSWER:</em> <em>NO + O ----> NO2 δ h = -232 kj</em>
