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iVinArrow [24]
3 years ago
15

Which of the following statements is not correct? 0.20 mole of O2 = 6.4 g 0.75 mole of H2CO3 = 47 g 3.42 moles CO = 95.8 g 4.1 m

oles Li2O = 94 g
Chemistry
2 answers:
Cerrena [4.2K]3 years ago
7 0
Its the value of co=95.8
aleksley [76]3 years ago
4 0

(D) 4.1 moles Li2O = 94g

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What is the maximum number of moles of Al2O3 that can be produced by the reaction of .4 mol of Al with .4 mol of O2
Darya [45]

<u>Given:</u>

Moles of Al = 0.4

Moles of O2 = 0.4

<u>To determine:</u>

Moles of Al2O3 produced

<u>Explanation:</u>

4Al + 3O2 → 2Al2O3

Based on the reaction stoichiometry:

4 moles of Al produces 2 moles of Al2O3

Therefore, 0.4 moles of Al will produce:

0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3

Similarly;

3 moles O2 produces 2 moles Al2O3

0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles

Thus Al will be the limiting reactant.

Ans: Maximum moles of Al2O3 = 0.2 moles

4 0
3 years ago
Read 2 more answers
Help. If an element has a charge of 4+, how many more protons does it have than electrons?
Andreyy89
Charge = Number of Protons - Number of Electrons = 4

So, Your Answer would be: 4

Hope this helps!
8 0
3 years ago
Why does the solution decolorize on standing after the equivalence point has been reached?
Arisa [49]

The equivalence point of a titration is equal to its stoichiometric equivalents of analyte and titrant.

Depending on the concentration of titrant we could be adding little excess of it and this may result in persistence of color of solution. After continuous stirring for a while the excess titrant may react with dissolved CO₂ in air and thus decolorizing the solution.

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8 0
4 years ago
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For the gas phase decomposition of t-butyl propionate,C2H5COOC(CH3)3 (CH3)2C=CH2 + C2H5COOHthe rate constant at 540 K is 8.90×10
kvv77 [185]

Answer:

The activation energy is 164.02 kJ/mol

Explanation:

Log (k2/k1) = Ea/2.303R × [1/T1 - 1/T2]

k1 = 8.9×10^-4 s^-1

k2 = 9.83×10^-3 s^-1

R = 8.314 J/mol.K

T1 = 540 K

T2 = 578 K

Log (9.83×10^-3/8.9×10^-4) = Ea/2.303×8.314 × [1/540 - 1/578]

1.043 = 6.359×10^-6Ea

Ea = 1.043/6.359×10^-6 = 164020 J/mol = 164020/1000 = 164.02 kJ/mol

6 0
3 years ago
One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc
Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

\text{The new partial pressure for }N_2 \ gas}

P_1V_1=P_2V_2

P_2=\dfrac{P_1V_1}{V_2} \\ \\  P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar

\text{The new partial pressure for }Ar \ gas}

P_2=\dfrac{P_1V_1}{V_2} \\ \\  P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar

Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \  \ \  \ \  \ \ \ \   \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \  \ \  \ \  \ \ \ \   \ \ \ \ \ \ \ \ = 2.225 \ Bar

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

2.1  \ bar = 2.07  \ atm \\ \\3.4 \  bar = 3.36 \  atm

For moles N₂:

PV = nRT \\ \\  n = \dfrac{PV}{RT}

n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}

n = 0.08297 \ mol  \ N_2

For moles of Ar:

PV = nRT \\ \\  n = \dfrac{PV}{RT}

n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}

n = 0.2172 \ mol  \ Ar

\mathtt{total \  moles = moles \ of \  N_2 + moles  \ of \ Ar}

=0.08297 mol + 0.2037 mol \\                   = 0.2867 mol gases

Finally;

The final pressure of the mixture is:

PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}

P = 2.217 atm

P ≅ 2.24 bar

7 0
3 years ago
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