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3 years ago

3. A representation of one unit of KCl in water is shown below. (The water molecules are intentionally not shown.)

Chemistry

1 answer:

alukav5142 [94]3 years ago

7 0

The representation is showing potassium atom (K) and Chlorine atom (Cl) when it ought to show their ions i.e potassium ion (K⁺) and Chlorine ion (Cl¯)

<h3>Dissociation equation for KCl</h3>

When potassium chloride, KCl dissolves in water, it dissociate to produce potassium ion (K⁺) and Chlorine ion (Cl¯) as shown below:

KCl(aq) —> K⁺(aq) + Cl¯(aq)

The representation given in the question is only showing potassium atom (K) and Chlorine atom (Cl). This makes it wrong as dissolution of ionic compounds in water will results in the corresponding ions of the element that makes up the compound

Please see attached photo

Learn more about dissociation equation:

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Even though so much energy is required to form a metal cation with a 2+ charge, the alkaline earth metals form halides with gene

nasty-shy [4]

The ∆Hrxn of the reaction is -394.5kJ/mol.

<h3>What is alkaline earth metal? </h3>

The alkaline earth metals are those elements which correspond to group 2 of the modern periodic table.

All elements of this group forms a cation of +2 charge.

The other elements of this group are:

Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium.

<h3>What is Halogen? </h3>

The Halogen elements are present in group 17 of the modern periodic table.

All elements of this groups forms anions of -1 charge.

The elements of this group are:

Fluorine, Chlorine, bromine, iodine, and astatine.

∆Hrxn = ∆H(bond broken) - ∆H(bond formed)

We have following data of bond energy in kJ/mol:

Mg—Mg = 738

Cl—Cl = -349

Mg—Cl = 783.5

Since, one mole of Mg react with one mole of Cl atom to form one mole of MgCl

∆Hrxn = 738-349-783.5

∆Hrxn = -394.5kJ/mol.

Thus, we concluded that the ∆Hrxn of the reaction is -394.5kJ/mol.

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2 years ago

When you slide a box across the floor what force must your push be stronger than?

weeeeeb [17]

If I remember correctly, the answer is <span>Friction force.</span>

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3 years ago

When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H

algol [13]

Answer: The empirical formula is $CH_2$ and molecular formula is $C_4H_8$

Explanation:

We are given:

Mass of $CO_2$ = 18.95 g

Mass of $H_2O$ = 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, = $\frac{12}{44}\times 18.59=5.07g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, = $\frac{2}{18}\times 7.759=0.862g$ of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.422}{0.422}=1$

For H = $\frac{0.862}{0.422}=2$

The ratio of C : H = 1: 2

Hence the empirical formula is $CH_2$ .

The empirical weight of $CH_2$ = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4$

The molecular formula will be= $4\times CH_2=C_4H_8$

6 0

3 years ago

1.what is the mass of an atom.<br>2.what is the difference between an ion and atom.

blsea [12.9K]

Answer: Atoms are single neutral particles, and an ion is a positively or negatively charged particle.

Explanation:

6 0

3 years ago

Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f

pishuonlain [190]

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>

<em />

<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>

T₂ = 0.5 T₁. According to Gay-Lussac's law,

$\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}$

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

$P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}$

P₂ in terms of the ideal gas equation is:

$P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}$

3 0

3 years ago