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3 years ago

Which resource produces the cleanest energy? coal gasoline oil wind

Chemistry

2 answers:

Alja [10]3 years ago

8 0

Answer: Wild is the cleanest one

stealth61 [152]3 years ago

4 0

Answer:

wind

Explanation:

wind is a natural and renewable resource

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What is the theoretical yield of aluminum oxide if 3.00 mol of aluminum metal is exposed to 2.55 mol of oxygen?

lisabon 2012 [21]

Theoretical yield of Al₂O₃: 1.50 mol.

<h3>Explanation</h3>

$2 \; \text{Al} + \dfrac{3}{2} \; \text{O}_2 \to {\bf 1} \; \text{Al}_2\text{O}_3$ ;

$4 \; \text{Al} + 3 \; \text{O}_2 \to 2 \; \text{Al}_2\text{O}_3 \; \textit{Balanced}$ .

How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?

Aluminum reacts to aluminum oxide at a two-to-one ratio.

$3.00 \times \dfrac{1}{2} = 1.50 \; \text{mol}$ .

As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?

Oxygen reacts to produce aluminum oxide at a three-to-two ratio.

$2.55 \times \dfrac{2}{3} = 1.70 \; \text{mol}$

As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced?

Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.

4 0

3 years ago

You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f

dsp73

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

( $1 atm = 760 mmHg$ )

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65 K

Putting values in above equation, we get:

$(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole$

Percentage recovery of carbon dioxide gas = 49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

$\frac{49.4}{100}\times 0.0438 mol=0.02164 mol$

$2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2$

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

$\frac{2}{1}\times 0.02164 mol=0.04328 mol$

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

5 0

4 years ago

What volume of concentrated (10.2 M) HCl would be required to prepare 1.11 x 104 mL of 1.5 M HC1? Enter your answer in scientifi

Tomtit [17]

Answer:

The required volume is 1.6 x 10³mL.

Explanation:

When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:

C₁ . V₁ = C₂ . V₂

where,

C₁ and V₁ are the concentration and volume of the concentrated solution

C₂ and V₂ are the concentration and volume of the dilute solution

In this case, we want to find out V₁:

C₁ . V₁ = C₂ . V₂

$V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL$

3 0

3 years ago

Give the coordination number of the central metal ion in the following complex ions.

melomori [17]

Answer:

a). Coordination no. of $[Ni(en)_2(NH_3)_2]^2+$ = 6

b). Coordination no. of $[Fe(en)(ox)Cl_2]^2-$ = 6

Explanation:

Coordination number is defined as number of donar atoms bonded to the central atom of the complex ion.

a). Coordination no. of $[Ni(en)_2(NH_3)_2]^2+$ = 6

en or ethylenediamine is a bidentate ligand.

In bidentate ligand, two atoms directly bonded to the central atom.

NH3 is a unidentate ligand.

So, coordination no.= No. of bidentate ligand x 2 + No. of unidentate ligand

= $2\times 2 + 2 = 6$

b). Coordination no. of $[Fe(en)(ox)Cl_2]^2-$ = 6

Ethylenediamine (en) is a bidentate ligand.

oxalate ion (ox) is also a bidentate ligand.

Cl is a unidentate ligand

So, coordination no.= No. of bidentate ligand x 2 + No. of unidentate ligand

= $2\times 2 + 2 = 6$

8 0

3 years ago

Is Boyle's law universally true? if not mention one of the its limitations

murzikaleks [220]

Yes because look in the book dh

4 0

2 years ago