Ill give brainlest to the one who answers right.

A gas with 4.0 atmospheres of pressure has a temperature of 27°C.

Gala2k [10]

Answer:

7.462

Explanation:

Well, every time that the tempurature is increased, the atmspheric pressure is increased by 0.574%. This would then mean that you would have 0.574 times

13. That would then equal 7.462. I hope this helps.

6 0

2 years ago

Which substance has Hf defined as 0 kJ/mol? H2O (s) Ne (l) O2 (g) CO2 (g)

vlabodo [156]

Answer:

O₂ (g).

Explanation:

All elements that found in nature has heat of formation (Hf) equal 0 J/mol.

So, the substance that has heat of formation (Hf) equal 0 J/mol is O₂ (g).

H₂O is formed from the reaction of hydrogen and oxygen and has Hf, Also, the solidification of H₂O (s) from H₂O (l) has (Hf).

Although, Ne (g) is found in nature and has no Hf, the liquification to form Ne (l) has heat of formation (Hf).

CO₂ (g) is formed from the reaction between carbon and oxygen and so has heat of formation (Hf).

8 0

3 years ago

Read 2 more answers

What orbital is represented by the transition metals in period four

andreev551 [17]

The name transition metal refers to the position in the periodic table of elements. The transition elements represent the successive addition of electrons to the d atomic orbitals of the atoms. In this way, the transition metals represent the transition between group 2 (2A) elements and group 13 (3A) elements.

8 0

3 years ago

In a titration, 25.0 mL of 0.500 M KHP is titrated to the equivalence point with NaOH. The final solution volume is 53.5 mL. Wha

Lena [83]

M1v1=m2v2
m2=(m1v1)/v2
Where m is the molarities and v is the volumes
<span>m2=(25.0*0.500)/53.5
m2=12.5/53.5
m2=0.2336
by rounding off:
m2=0.234 M
so the answer is C: 0.234 M</span>

3 0

3 years ago

Read 2 more answers

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