All categories

3 years ago

For the chemical reaction hclo4(aq)+naoh(aq)⟶h2o(l)+naclo4(aq) write the net ionic equation, including the phases.

1 answer:

5 0

The net ionic equation is as below

H^+(aq) + H^-(aq) → H2O(l)

Explanation
from the chemical equation

HClO4(aq) + NaOH (aq) → H2O(l) + NaClO4(aq)

write ionic equation
that is,
H^+(aq)+ClO4^-(aq) + Na^+(aq) +OH^-(aq)→ H2O(l) + Na^+(aq)+ClO4^-(aq)

cancel the spectator ions in both side(ions that do no take place in a chemical equation)
that ClO4^- and Na^+

the net ionic is therefore
= H^+(aq) + OH^-(aq) → H2O(l)

You might be interested in

Part C<br> Number of molecules in 8.437x10-2 mol C6H6

N76 [4]

Answer:

There are $5.08\times 10^{22}\ \text{molecules of}\ C_6H_6$

Explanation:

In this problem, we need to find the number of molecules in $8.437\times 10^{-2}$ mol of $C_6H_6$ .

The molar mass of $C_6H_6$ is $6\times 12+1\times 6=78\ g/mol$

No of moles = mass/molar mass

We can find mass from above formula.

$m=n\times M\\\\m=8.437\times 10^{-2}\ mol\times 78\ g/mol\\\\m=6.58\ g$

Also,

No of moles = no of molecules/Avogadro number

$N=n\times N_A\\\\N=8.437\times 10^{-2}\times 6.023\times 10^{23}\\\\N=5.08\times 10^{22}\ \text{molecules}$

Hence, there are $5.08\times 10^{22}\ \text{molecules of}\ C_6H_6$

5 0

3 years ago

The equilibrium constant Kc for the reaction below is 0.00584 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentra

Vinil7 [7]

Answer:

Explanation:

Given that:

The chemical equation for the reaction is:

Br2(g) ⇌ 2Br(g)

Initially 0.0345M 0.0416M

$Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016$

$Q_C =0.05016 >>> K_c(0.00584)$

Thus, the given reaction will proceed in the backward direction

The I.C.E table is as follows:

Br2(g) ⇌ 2Br(g)

I 0.0345 0.0416

C +x -2x

E (0.0345+x) (0.0416 -2x)

$K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584$

= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)

= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x

= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908 - 0.17184x + 4x²

Solving by using Quadratic formula

x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] = (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] = (0.0416 -2(0.0126)) = 0.0164 M

4 0

3 years ago

Ammonia, NH3 is a common base with Kb of 1.8 X 10-5. For a solution of 0.150 M NH3:

Vesnalui [34]

The concentrations : 0.15 M

pH=11.21

<h3>Further explanation</h3>

The ionization of ammonia in water :

NH₃+H₂O⇒NH₄OH

NH₃+H₂O⇒NH₄⁺ + OH⁻

The concentrations of all species present in the solution = 0.15 M

Kb=1.8 x 10⁻⁵

M=0.15

$\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}$

$\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21$

8 0

3 years ago

Chemical weathering processes do not alter the chemical structure of minerals in rocks. *

Anna007 [38]

I THINK that that is true. It's just weathering the rock down not change composition. Lemme know if you get it right please! (:

4 0

3 years ago

Sulfur reacts with fluorine to produce three different compounds. The mass ratio of fluorine to sulfur for each compound is give

NARA [144]

Answer:

Explanation:

Mass of F / Mass of S = 2.962/1 =2.962 X 32 / 32 = 94.78/32

Mass of F / Mass of S = 2.370 /1 = 2.370 X 32 / 32 = 75.84 /32

Mass of F /Mass of S = 3.555/1 = 3.555 x 32 / 32 = 113.76 / 32 .

Now constant mass of S that is 32 g reacts with different mass of F. They are as follows :

94.78 g , 75.84 g , and 113.76 g

Their ratio = 94.78 : 75.84 : 113.76

divide them by 19

their ratio = 5 : 4 : 6

So these data are consistent with law of multiple proportion.

8 0

3 years ago