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user100 [1]
3 years ago
10

17. When 1 mol of the following substances react completely with oxygen, the

Chemistry
2 answers:
Len [333]3 years ago
3 0
I would say Na. Oxygen has 2 valence electrons and when reacting with other molecules, the ones with the fullest or emptiest shells will react the least. Both H2 and Na are in the Alkali Metals in the first row, but since H2 has 2 molecules, it would use more oxygen than Ana
goldenfox [79]3 years ago
3 0

Answer:

Option A. Na

Explanation:

Let us generate a balanced equation for the reaction of O2 with the individual elements. This is illustrated below:

1. 4Na + O2 —> 2Na2O

From the equation,

4moles of Na comsumed 1mole of O2.

Therefore, 1mole of Na will comsume = 1/4 = 0.25mol of O2.

2. 2Ca + O2 —> 2CaO

From the equation,

2moles of Ca consumed 1 mole of O2.

Therefore, 1mole of Ca will consume = 1/2 = 0.5mole of O2

3. 4Al + 3O2 —> 2Al2O3

From the equation,

4moles of Al consumed 3moles of O2.

Therefore, 1mole of Al will consume = 3/4 = 0.75mol of O2

4. 2H2 + O2 —> 2H2O

From the equation,

2moles of H2 consumed 1mole of O2.

Therefore, 1mole of H2 will consume = 1/2 = 0.5mole of O2

The following data were obtained from the above calculations:

1mole of Na requires 0.25mol of O2

1mole of Ca requires 0.5mole of O2

1mole of Al requires 0.75mol of O2

1mole of H2 requires 0.5mole of O2

From the above, we see clearly that 1mole of Na consumes the least amount of O2

You might be interested in
PLEASE HELP
neonofarm [45]

Answer:

The answer to your question is 0.10 M

Explanation:

Data

Molarity = ?

mass of Sucrose = 125 g

volume = 3.5 l

Formula

Molarity = moles / volume

Process

1.- Calculate the molar mass of sucrose

C₁₂H₂₂O₁₁ = (12 x 12) + (1 x 22) + (16 x 11)

               = 144 + 22 + 176

               = 342 g

2.- Convert the mass of sucrose to moles

                  342 g of sucrose ------------------- 1 mol

                  125 g of sucrose -------------------- x

                           x = (125 x 1) / 342

                           x = 0.365 moles

3.- Calculate the molarity

Molarity = 0.365 / 3.5

4.- Result

Molarity = 0.10

5 0
3 years ago
A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0
3 years ago
Explain Charles law​
kodGreya [7K]

"Charles Law" is the relationship between "volume-temperature" (V ∝ T).

<u>Explanation</u>:  

Charles Law is basically volume-temperature relationship (V ∝ T). French scientist “Jacques Charles” in 1787 studied effect of temperature on the volume of gases at constant pressure, which described how gases manage to expand when heated. Law stated as “At constant pressure, the volume of a given "mass" of a gas decreases or increases by 1/273 of its volume at 0^{\circ} \mathrm{C} for each one degree rise or fall in temperature”. Formula derived from law is as follows: \mathbf{V}_{\mathbf{t}}=\mathbf{V}_{0}[\mathbf{1}+\mathbf{t} / \mathbf{2} 7 \mathbf{3}] here {V}_{0} is volume of given mass of a gas at 0^{\circ} \mathrm{C}, {V}_{t} is its volume at any temperature t^{\circ} \mathrm{C}. Application of Charles law is hot air balloons.

8 0
3 years ago
Which type of energy is generated due to the movement of charged particles
masya89 [10]

Answer:

electrical energy

Explanation:

just took the test

3 0
3 years ago
Read 2 more answers
A sample of an ideal gas has a volume of 3.25 L at 12.80 °C and 1.50 atm. What is the volume of the gas at 24.40 °C and
Bogdan [553]

Answer:

You have got the Combined Gas Equation for an Ideal Gas, which holds that  

P

1

V

1

T

1

=

P

2

V

2

T

2

V

2

=

P

1

×

V

1

×

T

2

T

1

×

P

2

...

Explanation:

5 0
3 years ago
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