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2 years ago

What is the order of the stars from Hottest to Coldest?

Chemistry

2 answers:

klio [65]2 years ago

6 0

Answer:

It's blue, white, yellow, orange, red

Explanation:

All of the stars are in different classes from O to M Blue is the hottest temperature at O and Red is the coolest temperature at M

O-Blue

B-Blue white

A-White

F-Yellowish White

G-yellow

K-Orange

M-Red

In your case it would be O,A,G,K,M

Hope that helped :)

Nikitich [7]2 years ago

4 0

Blue, white, yellow, orange, red

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Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

2 years ago

Which of the following can form a hydrogen bond with the HF molecule?

Lostsunrise [7]

Answer:

helium

(He)

Explanation:

helium

(He)

6 0

2 years ago

The entropy change for a real, irreversible process is equal to:______.

taurus [48]

Answer:

The entropy change for a real, irreversible process is equal to zero.

The correct option is 'c'.

Explanation:

Lets look around all the given options -:

(a) the entropy change for a theoretical reversible process with the same initial and final states , since the entropy change is equal and opposite in reversible process , thus this option in not correct.

(b) equal to the entropy change for the same process performed reversibly ONLY if the process can be reversed at all. Since , the change is same as well as opposite too . Therefore , this statement is also not true .

(c) zero. This option is true because We generate more entropy in an irreversible process. Because no heat moves into or out of the surroundings during the procedure, the entropy change of the surroundings is zero.

(d) impossible to tell. This option is invalid , thus incorrect .

Hence , the correct option is 'c' that is zero.

8 0

3 years ago

The absorbance of a garbanzo bean solution that had been diluted by a factor of three was 0.528. what was the concentration of t

Serjik [45]

The Beer-Lambert law states that A = E*c*l where A is absorbance, E is the molar absorbance coeffecient, c is concentration and l is path length. Therefore the absorbance is directly proportional to concentration, and by increasing the concentration by a factor of 3, absorbance will increase by a factor of 3 giving A = 1.584

6 0

3 years ago

Wsphorus-SL 3. In the isotope B-11, what does the 11 represent?

gregori [183]

Answer:

Here boron-11 means the name of the element is boron and the mass number is 11

Explanation:

8 0

2 years ago