Which statement best describes what a scientist is

Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze

Rashid [163]

Answer:

a. Δ $H^0_{rxn} = -108.0\frac{kJ}{mol}$

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate Δ $H^0_{rxn}$ , for the gas-phase reaction ) using the Hess's Law.

Δ $H^0_{rxn}$ = $E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}$

Given from the question, the table below shows the corresponding Δ $H^0_{t}(kJ/mol)$ for each compound.

Compound $H^0_{t}(kJ/mol)$

Liquid EO -77.4

$CH_4_(g_)$ -74.9

$CO_(g_)$ -110.5

If we incorporate our data into the above previous equation; we have:

Δ $H^0_{rxn}$ = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

= $-108.0 \frac{kJ}{mol}$

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

$T_{initial}$ = 93.0°C &

the enthalpy of vaporization (Δ $H^0_{vap}$ ) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as = $\frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}$

Let's not forget as well, that Δ $H^0_{vap}$ = $\frac{q}{mass}$

If we substitute Δ $H^0_{vap}$ for $\frac{q}{mass}$ in the above equation, we have;

specific heat capacity (c) = $\frac{deltaH^0_{vap}}{T_{final}-T_{initial}}$

Making ( $T_{final}- T_{initial}$ ) the subject of the formula; we have:

$T_{final}- T_{initial}$ = $\frac{delat H^0_{vap}}{specificheat capacity}$

$(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}$

$T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C$

= 227.76°C +93.0°C

= 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C

7 0

3 years ago

Hi, someone knows how to balance the following equation, and if you can thank you if you can explain how that is done:

kari74 [83]

To begin with, the equation given is not correct.
Correct equation is : CaCO3 + HCl ---> CaCl2 + H2O + CO2
It's CaCl2 not CaCl because Ca has a valency of 2

LHS RHS
CaCO3 + HCl ---> CaCl2 + H2O + CO2
First of all, to balance the equation you must look at the number of atoms on each side of the equation.
we have 2 H on the RHS and 1 H on the LHS. So, we put a 2 on the LHS

CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
Check for the LHS: 1 Ca, 1 C, 3 O, 2 H & 2 Cl on the LHS
Now check for the RHS: 1 Ca, 2 Cl, 2 H, 1 C & 3 O

Hope it helped!

7 0

3 years ago

Find the atomic numbers of the as yet undiscovered next two members of the series. radon

denpristay [2]

This answer is based on the electron configuration.

And you can use Aufbau's rule to predict the atomic number of the next elements.

Radon, Rn is the element number 86.

Following Aufbau's rules, the electron configuration of Rn is: [Xe] 6s2 4f14 5d10 6p6. This means that you are suming 2 + 14 + 10 + 6 = 32 electrons with respect to the element Xe.

You can verity that the atomic number of Xe is 54, so when you add 32 you get 54 + 32 = 86, which is the atomic number of Rn.

Again, as per Aufbau's rules, the next element of the same group or period is when the 6 electrons of the 7p orbital are filled. For that, they have to pass 32 elements whose orbitals are:

7s2 5f14 6d10 7p6: count the electrons added: 2 + 14 + 10 + 6 = 32, and that is why the next element wil have atomic number 86 + 32 = 118.

Now, when you go for a new series, you find a new type of orbital, the g orbital, for which the model predict there are 18 electrons to fill.

So the next element of the group will have ; 2 + 18 + 14 + 10 + 6 = 50 electrons, which means that the atomic number of this, not yet discovered element, has atomic number 118 + 50 = 168.

By the way the element with atomic number 118 was already discovdered at its symbol is Og. You can search that information in internet.

Answers: 118 and 168

7 0

3 years ago

Read 2 more answers

Predict the boiling point of water at a pressure of 1.5 atm.

Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

$ln\frac{P_{1} }{P_{2}} =$ -Δ $\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )$

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

$T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}$

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

7 0

3 years ago

4. How many J of energy are needed to raise the temperature of 165 g of water from 10.55°C to 47.32°C?

jeka94

Answer:

Q = 25360.269 j

Explanation:

Given data:

Mass = 165 g

Initial temperature = 10.55 °C

Final temperature = 47.32°C

Energy absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 47.32°C - 10.55 °C

ΔT = 36.77 °C

Q = m.c. ΔT

Q = 165 g . 4.18 j/g.°C . 36.77 °C

Q = 25360.269 j

3 0

3 years ago