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3 years ago

The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 Â°c is 4.48. what is the value of ka for hbro?

1 answer:

6 0

Hbro dissociate as follows
HBro---> H+ + BrO-
Ka= (H+)(BrO-) / HBro
PH = -log (H+)
therefore (H+) = 10^-4.48= 3.31 x10^-5
ka is therefore= ( 3.31 x 10^-5)^2/0.55=1.99 x10^-9

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Why does an atom of 23NA have 23 electrons

Julli [10]

It is actually something standardized more so than theoretical, however in terms of atoms in general the electrons are based on the amount of electronic shells that an atom has and the amount of electrons that atom can accommodate. In the case of sodium, it has three shells with 2 electrons on the first shell which is the maximum, 8 on the second shell which is also the max and 1 on their final shell
So simply put an ATOM of Sodium (Na) has 23 electrons because it has 23 protons which is a fact
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3 years ago

The mass of a deuterium nucleus 21H is less than its components masses. Calculate the mass defect.________ amu

Inessa [10]

Answer:

The mass defect of a deuterium nucleus is 0.001848 amu.

Explanation:

The deuterium is:

$^{A}_{Z}X \rightarrow ^{2}_{1}H$

The mass defect can be calculated by using the following equation:

$\Delta m = [Zm_{p} + (A - Z)m_{n}] - m_{a}$

Where:

Z: is the number of protons = 1

A: is the mass number = 2

$m_{p}$ : is the proton's mass = 1.00728 amu

$m_{n}$ : is the neutron's mass = 1.00867 amu

$m_{a}$ : is the mass of deuterium = 2.01410178 amu

Then, the mass defect is:

$\Delta m = [1.00728 amu + (2- 1)1.00867 amu] - 2.01410178 amu = 0.001848 amu$

Therefore, the mass defect of a deuterium nucleus is 0.001848 amu.

I hope it helps you!

5 0

3 years ago

Read 2 more answers

Write the oxidation and reduction half reactions;

luda_lava [24]

Answer:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

$Br_2+2e^-$ ⇒ $2Br^-$

$Mg$ ⇒ $Mg^{2+}+2e^-$

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

Explanation:

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

$Br_2+2e^-$ ⇒ $2Br^-$

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

$Mg$ ⇒ $Mg^{2+}+2e^-$

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

3 0

3 years ago

Read 2 more answers

d) A short ton of coal holds about 2.24 x 107 J of energy and costs about $56.45 per short ton. How many Joules of energy in the

Feliz [49]

Answer:

396811.337 J

Explanation:

The cost of one short ton of coal = $56.45

The energy related to the short ton of coal = $2.24\times 10^{7}\ J$

Thus, As according to the question,

$56.45 of coal have $2.24\times 10^{7}\ J$ of energy.

$1 of coal have $\frac{2.24\times 10^{7}}{56.45}\ J$ of energy.

<u>The amount of energy = 396811.337 J</u>

4 0

3 years ago

the reaction of aluminum with chlorine gas is shown 2Al + 3Cl2 -> 2AlCl3 based on this equation how many molecules of chlorin

Vanyuwa [196]

45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

The balanced equation for the reaction is given below:

2Al + 3Cl₂ —> 2AlCl₃

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

Therefore,

30 atoms of Al will require = $\frac{30 * 3}{2}\\$ = 45 molecules of Cl₂.

Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

Learn more: brainly.com/question/24918379

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2 years ago