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Sedbober [7]
3 years ago
8

One molecule of the oxide of element Z reacts with six molecules of water to

Chemistry
1 answer:
zalisa [80]3 years ago
8 0
I can guess that Z is Phosphorus. So the oxide - P4O10 (but the most known its empirical formula P2O<span>5)</span>. Reaction with water:
P4O10 + 6H2O = 4H3PO4. Acidic compound is phosphoric acid.
You might be interested in
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
Find the weight of HNO_3 present in 20ml, 0.30 N
yanalaym [24]

Answer:

mass of HNO₃ = 0.378 g

Explanation:

Normality = Molarity * number of equivalents

Molarity = Normality/number of equivalents

normality of HNO₃ = 0.30 N, Volume = 20 mL

HNO₃ ionizes in the following way:

HNO₃(aq) ----> H⁺ + NO₃⁻

Therefore, number of equivalents for HNO₃ is 1

molarity of HNO₃ = 0.30/1 =0.30 mol/dm³

Using the formula, molarity = number of moles/volume in liters

number of moles = molarity * volume

Number of moles of HNO₃ = 0.30 mol/dm³ * 20ml * 1 dm³ /1000 mL

number of moles = 0.006 moles

From the formula, mass = number of moles * molar mass

molar mass of HNO₃ = 63.0 g/mol

mass = 0.006 * 63

mass of HNO₃ = 0.378 g

6 0
3 years ago
What major gases are absorb by the ocean
zysi [14]

Answer:

C) carbon dioxide and hydrigen

7 0
3 years ago
Which of the following frequencies corresponds to light with the longest wavelength? A) 3.00 times 10^13 s^-1 B) 4.12 times 10^5
Vsevolod [243]

Answer:

1) B. 4.12 times 10^5 s^-1

2) B. frequency-v

3) C. 1.18 times 10^15 s^-1

4) B. 4.39 times 10^-19 J

5) B. Energy is absorbed

6) C. 3

7) 1s^2  2s^2  2p^6  3s^2  3p^6  4s^2  3d^10  4p^6  5s^2  4d^10  5p^6     6s^2  

Explanation:

1)

The wavelength is inversely proportional to the frequency. Thus, the smallest frequency shall correspond to the longest wavelength. Thus, the correct answer is <u>B. 4.12 times 10^5 s^-1</u>

2)

<u>B. frequency-v</u> are wrongly paired because frequency is represented by f. Thus, correct pair will be frequency-f.

3)

The relationship between wavelength (λ) and frequency (f) is:

c = fλ

f = c/λ

where, c = speed of light = 3 x 10^8 m/s

f = (3x10^8 m/s) / (254 x 10^-9 m)

f = 1.18 x 10^15 s^-1

Thus, the correct option is <u>C. 1.18 times 10^15 s^-1.</u>

4)

The energy of photon is given as:

E = hc/λ

where, c = speed of light = 3 x 10^8 m/s

            h = Plank's Constant = 6.625 x 10^-34 J.s

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s) / (453 x 10^-9 m)

E = 4.39 x 10^-19 J

Thus, the correct option is <u>B. 4.39 times 10^-19 J</u>.

5)

Since, the higher energy levels away from nucleus have higher energies. So, in order to move an electron from lower to higher or distant energy level, it must absorb energy from external source. So, the correct option is <u>B. Energy is absorbed</u>.

6)

The P-Sub-level has three orbitals, each having two electrons. Thus, P-Sub-Level accommodates total of 6 electrons. The correct option is <u>C. 3</u>

7)

The electronic configuration of barium atom is:

<u>1s^2  2s^2  2p^6  3s^2  3p^6  4s^2  3d^10  4p^6  5s^2  4d^10  5p^6  6s^2  </u>

8 0
3 years ago
The earth rotates on an imaginary line called an?
Ierofanga [76]
The Earth rotates on an imaginary line called an axis, which runs through Earth from the North Pole to the South Pole. The rotation of the Earth causes day
8 0
3 years ago
Read 2 more answers
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