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3 years ago

What are the concentrations of h3o+ and oh− in tomatoes that have a ph of 4.10?

Chemistry

1 answer:

IRINA_888 [86]3 years ago

5 0

Answer : The concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

Solution : Given,

pH = 4.10

pH : pH is defined as the negative logarithm of hydronium ion concentration.

Formula used : $pH=-log[H_3O^+]$

First we have to calculate the hydronium ion concentration by using pH formula.

$4.10=-log[H_3O^+]$

$[H_3O^+]=antilog(-4.10)$

$[H_3O^+]=7.94\times 10^{-5}$

Now we have to calculate the pOH.

As we know, $pH+pOH=14$

$4.10+pOH=14$

$pOH=9.9$

Now we have to calculate the hydroxide ion concentration.

$pOH=-log[OH^-]$

$9.9=-log[OH^-]$

$[OH^-]=antilog(-9.9)$

$[OH^-]=1.258\times 10^{-10}$

Therefore, the concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

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Explain Rutherford Bohr model?

Liula [17]

Answer:

heya!!!

Explanation:

In atomic physics, the Rutherford–Bohr model or Bohr model, presented by Niels Bohr and Ernest Rutherford in 1913, is a system consisting of a small, dense nucleus surrounded by orbiting electrons—similar to the structure of the Solar System, but with attraction provided by electrostatic forces in place of gravity.

7 0

3 years ago

Given the balanced ionic equation representing the reaction in an operating voltaic cell: zn(s) + cu2+(aq) → zn2+(aq) + cu(s) th

soldi70 [24.7K]

Answer: from the Zn anode to the Cu cathode

Justification:

1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)

2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).

3) Then, you can already tell that electrons go from Zn to Cu.

4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.

So you get that the electrons flow from the anode (Zn) to the cathode (Cu).

Always oxidation occurs at the anode, and reduction occurs at the cathode.

3 0

3 years ago

Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.

JulsSmile [24]

Answer:

The volume of hydrogen gas produced will be approximately 50.7 liters under STP.

Explanation:

Relative atomic mass data from a modern periodic table:

H: 1.008;
Cl: 35.45.

Magnesium is a reactive metal. It reacts with hydrochloric acid to produce

Hydrogen gas $\rm H_2$ , and
Magnesium chloride, which is a salt.

The chemical equation will be something like

$\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}]$ ,

where the coefficients and the formula of the salt are to be found.

To determine the number of moles of $\rm H_2$ that will be produced, first find the formula of the salt, magnesium chloride.

Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.

On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be $\rm MgCl_2$ .

$\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq)$ .

Balance the equation. $\rm MgCl_2$ contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.

$\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq)$ .

The number of $\rm Mg$ and $\rm Cl$ atoms shall be the same on both sides. Therefore

$\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq)$ .

The number of $\rm H$ atoms shall also conserve. Hence the equation:

$\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq)$ .

How many moles of HCl are available?

$M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}$ .

$\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol$ .

How many moles of Hydrogen gas will be produced?

Refer to the balanced chemical equation, the coefficient in front of $\rm HCl$ is 2 while the coefficient in front of $\rm H_2$ is 1. In other words, it will take two moles of $\rm HCl$ to produce one mole of $\rm H_2$ . $\rm 4.52576\;mol$ of $\rm HCl$ will produce only one half as much $\rm H_2$ .

Alternatively, consider the ratio between the coefficient in front of $\rm H_2$ and $\rm HCl$ is:

$\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}$ .

$\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol$ .

What will be the volume of that many hydrogen gas?

One mole of an ideal gas occupies a volume of 22.4 liters under STP (where the pressure is 1 atm.) On certain textbook where STP is defined as $\rm 1.00\times 10^{5}\;Pa$ , that volume will be 22.7 liters.

$V(\text{H}_2) = \rm 2.26288\;mol\times 22.4\;L\cdot mol^{-1} = 50.69\; L$ , or

$V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L$ .

The value "165.0 grams" from the question comes with four significant figures. Keep more significant figures than that in calculations. Round the final result to four significant figures.

5 0

3 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

3 years ago

Which of the following groups in the periodic

Dimas [21]

Elements of Group 1 and group 2 in the periodic table contain elements so reactive that they are never found in the free state

<u>Explanation</u>:

The metals in group 1 of periodic table consisting of 'alkali metals' which include lithium, potassium, sodium, rubidium, Francium and caesium. They are highly reactive because they have low ionisation energy and larger radius. The group 2 metals consist of 'alkaline earth metals' which include calcium, strontium, barium, beryllium, radium and magnesium. These alkaline earth metal have +2 oxidation number, hence are highly reactive.

These both group metals are mostly reactive and so are never found in a free state. When they are exposed to air they would immediately react with oxygen. Hence, are stored in oils to avoid oxidation.

7 0

3 years ago