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2 years ago

What is the Arrhenius definition of an acid

Chemistry

2 answers:

matrenka [14]2 years ago

8 0

The answer to your question is,

An Arrhenius acid is a type of substance that separates in water to form Hydrogen ions.

-Mabel <3

Ugo [173]2 years ago

6 0

Explanation:

According to Arrhenius concept :

An acid is defined as a substance which donates hydronium ions $(H_3O^+)$ in water. For example hydrochloric acid, sulfuric acid. etc.

$HA(aq)+H_2O\rightarrow A^-(aq)+H_3O^+(aq)$

A base is defined as a substance which donates hydroxide ions $(OH^-)$ when dissolved in water .For example : sodium hydroxide , calcium hydroxide etc.

$BOH(aq)\rightarrow B^+(aq)+OH^-(aq)$

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Name the following alkyne: CH3CH₂C = CCH₂CH₂CH3

liubo4ka [24]

The given alkyne is Option A 3-heptyne

<h3>What is an Alkyne ?</h3>

The hydrocarbon having at least one C-C triple bond is called an Alkyne.

It has the general formula of $\rm C_n H_{2n+2}$ .

In the question it is being mentioned that it is an alkyne so there will be a triple bond and not a double bond.

It has been asked in the question that

CH3CH₂C ≡ CCH₂CH₂CH3 is which alkyne from the given option.

The counting of the Carbon chain is done from the left side and the Triple bond is at the 3rd Carbon , so 3-heptyne .

To know more about Alkyne

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6 0

1 year ago

2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an

Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0

2 years ago

A) Brass Key

ruslelena [56]

B I’m pretty sure. (Sorry if I wrong-)

7 0

2 years ago

The H 2 produced in a chemical reaction is collected through water in a eudiometer. If the pressure in the eudiometer is 101.3 k

Gnom [1K]

The H2 produced in a chemical reaction is collected through water in a eudiometer. The pressure (kPa) of the H2 gas is 98.89 kPa

The total pressure in a chemical reaction is the total sum of the partial pressure and the vapor pressure of the chemical substances taking place in the chemical reaction.

Total pressure = partial pressure of H₂ gas + vapor pressure of H₂O

∴

The vapor pressure of H₂ gas = Total pressure in the eudiometer - partial pressure of H₂O

Given that:

The total pressure in the eudiometer = 101.3 kPa
The partial pressure of H₂O = 2.41 kPa

The vapor pressure of H₂ gas = 101.3 kPa - 2.41 kPa

The vapor pressure of H₂ gas = (101.3 - 2.41) kPa

The vapor pressure of H₂ = 98.89 kPa

Therefore, we can conclude that the vapor pressure of H₂ is 98.89 kPa.

Learn more about partial pressure here:

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3 0

1 year ago

A town is looking at multiple solutions for its beach erosion problem. Which factors of beach nourishment should the town consid

Art [367]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers