What two properties of a gas depend on its container

What is the maximum amount of kcl that can dissolve in 200 g of water? (the solubility of kcl is 34 g/100 g h2o at 20°c.) (1 poi

Solnce55 [7]

The answer is 68g
hope this helps!!

5 0

3 years ago

Which is the correct equation for the reaction of magnesium with hydrochloric acid to produce hydrogen gas and magnesium chlorid

djyliett [7]

Answer:

Mg(s) + 2 HCl(aq) --> MgCl 2(aq) + H 2(g)

8 0

2 years ago

Concentration of 10.00 mL of HBr if it takes 5 mL of a 0.253 M LiOH solution to<br> neutralize it?

Sonbull [250]

In a titration, for an acid to neutralize a base, at the equivalence point, there should be an equal number of moles of H+ and OH-.

Moles of OH- can be found by multiplying the concentration of the base by the volume. (You will need to keep in mind the stoichimetric coefficients if the strong base is Ca(OH)₂, Ba(OH)₂, or Sr(OH)₂.

Moles of OH- = moles of H+
(0.253 M) * 0.005 L = 0.01000 L * c
c = 0.1265 M

The concentration of HBr is 0.127 M.

3 0

3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago

The molar mass of H2O is 18.01 g/mol. The molar mass of O2 is 32.00 g/mol. What mass of H2O, ins grams, must react to produce 50

Ivan

Answer:

56.28 g

Explanation:

First change the grams of oxygen to moles.

(50.00 g)/(32.00 g/mol) = 1.5625 mol O₂

You have to use stoichiometry for the next part. Looking at the equation, you can see that for every 2 moles of H₂O, 1 mole of O₂ is produced. Convert from moles of O₂ to moles of H₂O using this relation.

(1.5625 mol O₂) × (2 mol H₂O/1 mol O₂) = 3.125 mol H₂O

Now convert moles of H₂O to grams.

(3.125 mol) × (18.01 g/mol) = 56.28125 g

Convert to significant figures.

56.28125 ≈ 56.28

5 0

3 years ago