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rosijanka [135]
1 year ago
15

What mass of AI2O3 forms, knowing the molar mass of Al2O3 is 102 g/mol ?

Chemistry
1 answer:
TiliK225 [7]1 year ago
3 0

Question

What mass of Al₂O₃ form is produced if 25.00 g of aluminum (al) is reacted completely?, knowing the molar mass of Al₂O₃ is 102 g/mol?

Atomic mass of Al =27 and O = 16

The molar mass of Al₂O₃ = 102.

Mass of aluminium in 102 g of Al₂O₃ = 54 g

54 g Al => 102 g Al₂O₃

25g Al => x g Al₂O₃

x = 25*102/54 = 47.2 g Al₂O₃

To solve more questions on molar mass check here:

brainly.com/question/21334167

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What happens to the energy added during a phase change?
Ivanshal [37]

Answer:You're answer is D: it is converted to kinetic energy

Explanation:

During a change of phase, the average kinetic energy of the molecules stays the same, but the average potential energy changes. ... My interpretation is that during a phase change, the temperature remains equal, but the kinetic energy of its particles increase/decrease.

                                             LINK:

https://chemistry.stackexchange.com/questions/82163/clarification-of-kinetic-energy-during-phase-change

7 0
3 years ago
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The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism
MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

5 0
3 years ago
1. How many protons does 14/6 C contain? What element is this?
Harman [31]

Answer:

It has 6 protons and its Carbon 14

Explanation:

3 0
3 years ago
a) Whatis the composition in mole fractions of a solution of benzene and toluene that has a vapor pressure of 35 torr at 20 °C?
iogann1982 [59]

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.  

For toluene and benzene would be:

P_{B}=x_{B}*P_{B}^{o}

P_{T}=x_{T}*P_{T}^{o}

Where:

P_{B} is partial pressure for benzene in the liquid  

x_{B} is benzene molar fraction in the liquid  

P_{B}^{o} vapor pressure for pure benzene.  

The total pressure in the solution is:

P= P_{T}+ P_{B}

And  

1=x_{B}+x_{T}

Working on the equation for total pressure we have:

P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}

Since x_{T}=1-x_{B}

P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}

We know P and both vapor pressures so we can clear x_{B} from the equation.

x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}

x_{B}=\frac{35- 22}{75-22} = 0.24

So  

x_{T}=1-0.24 = 0.76

To get the mole fraction for the vapor we know that in the equilibrium:

P_{B}=y_{B}*P

y_{T}=1-y_{B}

So  

y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}

y_{B}=\frac{0.24*75}{35}=0.51

y_{T}=1-0.51=0.49

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

7 0
3 years ago
Which compounds have the empirical formula CH2O?
lara [203]

Answer:

The answer to your question is all the formulas in bold has the same empirical formula

Explanation:

Data

Empirical formula     CH₂O

Process

To solve this problem factor the subscripts of each formula and compare the result with the empirical formula given.

a) C₂H₄O₂      factor 2          2(CH₂O)

b) C₃H₆O₃      factor 3          3(CH₂O)

c) CH₂O₂        this formula can not be simplified

d) C₅H₁₀O₅      factor 5         5(CH₂O)

e) C₆H₁₂O₆      factor 6         6(CH₂O)

8 0
3 years ago
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