Molarity: M = #moles of solute / liters of solution

# moles = mass / molar mass

Molar mass calculation

Barium hydroxide = Ba (OH)2

Atomic masses
Ba = 137.4 g/mol
O=16 g/mol
H=1 g/mol
Molar mass of Ba (OH)2 = 137.4 g/mol + 2*16g/mol + 2*1 g/mol = 171.4 g/mol

# mol = 25.0g/171.4 g/mol = 0.146 mol

For the volume of water use the fact that the density is 1g/ml., so 120 g = 120 ml = 0,120 liters.

M = 0.146mol / 0.120 liters = 1.22 mol/liter

3 0

4 years ago

The half-life for the reaction below was determined to be 2.14 × 10^4 s at 800 K. The value of the half-life is independent of t

lord [1]

Answer:

0.0907 s

Explanation:

This an Arrhenius equation problem, so you relate the half-life with the kinetic constant of the reaction in order to calcule the same thermodynamic parameters at another temperature.

To calcule the kinetic constant of the reaction you need to know the order of it, look closely to the sentence "The value of the half-life is independent of the inital concentration of N2O present." the only order independent from the initial concentration of reagents is first order, so you can calculate K at 800 K, using:

$k(800 K)=\frac{ln(2)}{t_{1/2}}= \frac{ln(2)}{2.14 * 10^{4} s}}=3.239*10^{-5}s^{-1}}$

Now you can use Arrhenius equation to calcule K at 1150.66 K

$ln(\frac{k1}{k2} )=-\frac{E_{a} }{R}(\frac{1}{T2} - \frac{1}{T1} )$

$k2= k1*exp(-\frac{E_{a} }{R}(\frac{1}{T2} - \frac{1}{T1} ))=3.239*10^{-5}s^{-1}*exp(-\frac{270 000 J/mol }{8.314 J/mol *k }(\frac{1}{1150.66K} - \frac{1}{800K} ))=7.639 s^{-1}$

Then calculate the new half-life:

$t_{1/2} =\frac{ln(2)}{7.639s^{-1}}=0.0907 s$

3 0

3 years ago

What happens to the molecule If energy is removed from a substance

Andrej [43]

Answer:

Removing energy (cooling) atoms and molecules decreases their motion, resulting in a decrease in temperature

7 0

2 years ago