<span>The equation you used is KE=hv-hv0, where h=6.63*10^-34 (constant). You multiply h by 1.5*10^15. Multiply h by the threshold freq of cesium (from part A). Subtract the second answer from the first answer, and you get the kinetic energy. Hope this helps.</span>

7 0

3 years ago

Dr. I. M. A. Brightguy adds 0.1727 g of an unknown gas to a 125-mL flask. If Dr. B finds the pressure to be 736 torr at 20.0°C,

AlladinOne [14]

Answer:

The gas that Dr. Brightguy added was O₂

Explanation:

Ideal Gases Law to solve this:

P . V = n . R . T

Firstly, let's convert 736 Torr in atm

736 Torr is atmospheric pressure = 1 atm

20°C = 273 + 20 = 293 T°K

125 mL = 0.125L

0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K

(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n

5.20x10⁻³ mol = n

mass / mol = molar mass

0.1727 g / 5.20x10⁻³ mol = 33.2 g/m

This molar mass corresponds nearly to O₂

7 0

3 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

3 years ago