The Mystery Salt
Imagine that you have a barrel of salt, but you forgot to label it. You know it must be either KNO3, or KCl.
You look at the solubility curves for KNO3 and KCl and you find that at 35 degrees Celsius, 100 g of water can dissolve about 30 g of KNO3, or about 37 g of KCl.
You have a scale, a hot plate, a thermometer, empty beakers, and plenty of water.
1. How can you determine whether the barrel contains KNO3 or KCl? Hint: What property could you use to identify the mystery salt? (1 point)
2. Explain exactly what you would do and why you would do it. (2 points)
3. How would your results identify the mystery salt? Hint: What results would indicate KNO3? What results would indicate KCl? (2 points)

The correct answer is option (A). The theoretical yield of hydrogen (H₂) is approximately 1.11 g.

To find the theoretical yield of hydrogen gas (H₂) produced in this reaction, follow these steps:

1. Determine the molar mass of HCl:

- The molar mass of HCl = 1.01 g/mol (for H) + 35.45 g/mol (for Cl) = 36.46 g/mol.

2. Calculate the number of moles of HCl in 40.0 g:

[tex]\text{moles of HCl} = \frac{40.0 \, \text{g}}{36.46 \, \text{g/mol}} = 1.095 \ moles[/tex]

3. Using the stoichiometry of the balanced equation:

For every 2 moles of HCl, 1 mole of H₂ is produced.

Therefore, moles of H₂ produced = [tex]\frac{1.095 \, \text{moles of HCl}}{2} = 0.5475 \ moles \ of \ H_2[/tex].

4. Determine the molar mass of H₂:

- The molar mass of H₂ = 2.016 g/mol.

5. Calculate the theoretical yield of H₂ in grams:

[tex]{mass\ of {H_2}} = 0.5475 \times 2.016 = 1.103824 g[/tex]

6. Round to the nearest gram: - 1.11 g

So, the theoretical yield of hydrogen gas is approximately: A) 1.11 g

The complete question is:

Consider the balanced equation. [tex]2HCl + Mg \rightarrow MgCl_2 + H_2[/tex] if 40.0 g of HCl react with an excess of magnesium metal, what is the theoretical yield of hydrogen?

A) 1.11 g

B) 2.22 g

C) 52.2 g 1

D) 104 g

The correctly balanced equation for the reaction between hydrogen (H2) and chlorine (Cl2) to form hydrogen chloride (HCl) is H2 + Cl2 → 2 HCl, with the coefficients being 1 for H2, 1 for Cl2, and 2 for HCl.

The question concerns balancing a chemical equation, which falls under the field of Chemistry, specifically with relation to stoichiometry, the part of chemistry that concerns the quantities of substances that come into play during chemical reactions.

In the chemical equation H2 + Cl2 → HCl, to balance it, you need to ensure that the number of atoms of each element is equal on both the reactant and product sides. After balancing the equation, it takes the form H2 + Cl2 → 2 HCl. This means that the coefficients for hydrogen (H2), chlorine (Cl2), and hydrogen chloride (HCl) are 1, 1, and 2 respectively.

Final answer:

To balance the equation between hydrogen gas (H₂) and chlorine gas (Cl₂) forming hydrogen chloride (HCl), we add a coefficient of 2 in front of HCl. This yields a balanced chemical equation where the coefficients for H₂, Cl₂, and HCl are 1, 1, and 2, respectively.

Explanation:

The equation given is a combination of hydrogen gas (H₂) and chlorine gas (Cl₂) to form hydrogen chloride (HCl). To balance the equation, we should have equal numbers of each type of atom on both sides of the equation. We start with hydrogen and chlorine atoms, and then balance the number of hydrogen chloride molecules produced.

Initially, we have the equation:

H₂ + Cl₂ → HCl

This is not balanced because we have 2 hydrogen atoms on the left and only 1 on the right; similarly, we have 2 chlorine atoms on the left but only 1 on the right. By placing a coefficient of 2 in front of HCl, which indicates we are producing 2 molecules of it, the equation becomes:

H₂ + Cl₂ → 2 HCl

Now, with two atoms of both hydrogen and chlorine on each side of the equation, the chemical equation is balanced. Therefore, the correct coefficients are 1 for H₂, 1 for Cl₂, and 2 for HCl, respectively.

1) polar solvents dissolve ionic or polar solutes

2) Non-polar solvents dissolve non-polar solutes

3) Non-polar solvents do not dissolve polar and ionic solutes 

4) Polar solvents do not dissolve non-polar solutes.

There are exceptions, but the rule is applicable in here.

Final answer:

The water has a Ka value, more precisely referred to as ionization constant (Kw), of 1.0 × 10^-14 at 25 °C.

Explanation:

The water has a Ka value that is actually known as the ionization constant for water, Kw. The Ka value is a specific term generally used for the acid dissociation constant of substances other than water. For water at 25 °C, the product of the concentrations of the hydrogen ions ([H3O+]) and the hydroxide ions ([OH-]) is 1.0 × 10^-14, so Kw is 1.0 × 10^-14. This means that in pure water, or in a neutral aqueous solution, the concentration of hydrogen ions and hydroxide ions are both 1.0 × 10^-7 M. Therefore, the correct answer to the question 'water has a Ka value of what?' is 1 × 10^-14.

The Ka and Kb reactions for the HSO3- ion in water, acting as an acid and base respectively, are: 1) Ka Reaction: HSO3-(aq) + H2O(l) → H3O+(aq) + SO3 2-(aq), 2) Kb Reaction: HSO3-(aq) + H2O(l) → OH-(aq) + H2SO3(aq).

The HSO3- ion is amphoteric, meaning it can act as both an acid and a base. The ka and kb reactions for this ion in water would be as follows:

In the Ka reaction, the bisulfite ion (HSO3-) donates a hydrogen ion (H+) to water, thereby acting as an acid. The resultant ions are hydronium (H3O+) and sulfite (SO3 2-).

In the Kb reaction, the bisulfite ion (HSO3-) accepts a hydrogen ion (H+) from water, thus acting as a base. The resulting species are hydroxide ion (OH-) and sulfurous acid (H2SO3).

https://brainly.com/question/15209937

#SPJ6

The balanced reactions for [tex]HSO_{3}^{-}[/tex]⁻ in water are [tex]HSO_{3}[/tex](aq) + H₂O(l) ⇌ [tex]SO^{2-} _{3}[/tex]⁻(aq) + H₃O⁺(aq) for its Ka, and [tex]HSO_{3} ^{-}[/tex]⁻(aq) + H₂O(l) ⇌ [tex]H_{2} SO_{3}[/tex](aq) + OH⁻(aq) for its [tex]k_{b}[/tex]. The equilibrium constant expressions for these reactions are written accordingly. This demonstrates [tex]HSO_{3} ^{-}[/tex] acting both as an acid and a base.

When  (hydrogen sulfite ion) behaves as an acid in water, it donates a proton (H⁺) to form [tex]SO_{3} ^{2-}[/tex]and H₃O⁺. The balanced equation for this equilibrium reaction is:

[tex]HSO_{3} ^{2-}[/tex](aq) + H₂O(l) ⇌ [tex]SO_{3} ^{2-}[/tex](aq) + H₃O⁺(aq)

The equilibrium constant expression for this reaction ([tex]k_{a}[/tex]) can be written as:

[tex]k_{a}[/tex] = [[tex]SO_{3} ^{2-}[/tex]⁻][H₃O⁺] / [[tex]HSO_{3} ^{-}[/tex]]

When [tex]HSO_{3} ^{-}[/tex] behaves as a base, it accepts a proton (H⁺) from water to form [tex]H_{2} SO_{3}[/tex] and OH⁻. The balanced equation for this equilibrium reaction is:

[tex]HSO_{3} ^{-}[/tex](aq) + H₂O(l) ⇌ [tex]H_{2} SO_{3}[/tex] (aq) + OH⁻(aq)

The equilibrium constant expression for this reaction ([tex]k_{b}[/tex]) can be written as:

Kb = [[tex]H_{2}SO_{3}[/tex]][[tex]OH_{-}[/tex]] / [[tex]HSO_{3} ^{-}[/tex]]

The empirical formula of the compound containing sodium, chlorine, and oxygen is 25.42% sodium by mass. A 3.25 g sample gives 4.33×10²² atoms of oxygen is NaClO₂.

Step 1: Calculate the mass of sodium in the sample

Step 2: Calculate the number of moles of sodium

Step 3: Calculate the number of moles of oxygen

Step 4: Calculate the mole ratio of sodium to oxygen

Step 5: Calculate the mass of chlorine in the sample

Step 6: Calculate the mole ratio of chlorine to sodium

Step 7: Write the empirical formula

Therefore, the empirical formula of the compound is NaClO₂.

Answer b is the answer for the equation:

Answer : The 10300 milliliter in [tex]cm^3[/tex] and L are, [tex]10300 cm^3[/tex] and 10.3 L respectively.

Explanation :

The conversion used from milliliters to centimeter cube is:

[tex]1ml=1cm^3[/tex]

As we are given the volume 10300 ml. Now we have to determine the volume in centimeter cube.

As, [tex]1ml=1cm^3[/tex]

So, [tex]10300ml=\frac{10300ml}{1ml}\times 1cm^3=10300cm^3[/tex]

The volume in centimeter cube is, [tex]10300cm^3[/tex]

The conversion used from milliliters to liter is:

1 ml = 0.001 L

As we are given the volume 10300 ml. Now we have to determine the volume in liter.

As, 1 ml volume = 0.001 L

So, 10300 ml volume = [tex]\frac{10300ml}{1ml}\times 0.001L=10.3L[/tex]

The volume in liter is, 10.3 L

Answer:

The 10300 milliliter in  and L are,  and 10.3 L respectively.

Explanation:

Yes, your answer is correct. Heating an air-filled, sealed can results in stronger and more frequent collisions of air molecules against the wall, leading to an increase in pressure.

When you heat an air-filled, sealed can, the collisions of air molecules against the wall of the can become stronger and more frequent. This is because as the temperature inside the can increases, the air molecules move faster and collide more energetically with the can walls. This increased movement results in a rise in pressure inside the can due to an increased number of collisions and force per collision. This concept is based on the principles of gas pressure and the kinetic molecular theory, which relate temperature, molecular speed, and pressure in a contained gas.

Gas pressure is indeed caused by collisions between gas molecules and the container walls. An increase in temperature causes the molecules to move faster, leading to more collisions with the walls, which translates into an increase in pressure. Furthermore, the concept that gas pressure can be increased by compressing a gas into a smaller volume explains why a canister feels cold when its gas is released: the surrounding air absorbs the energy from the expanding gas.

Answer:

3.3 atm

Explanation:

This is a simple application of Dalton's law of partial pressure which state that the total pressure exerted by a mixture of gas is the sum of the individual partial pressure of the component gases.

[tex]P_{total} = P_1 + P_2 + ........ + P_n[/tex]

Hence, the total pressure of the gas mixture becomes:

[tex]P_{total} = P_{methane} + P_{hydrogen} + ........ + P_{nitrogen}[/tex]

                         = 1 + 1.2 + 1.1

                                 = 3.3 atm

The total pressure of the mixture is 3.3 atm

If 23 g of sodium reacts completely with 71 g of chlorine, the limiting reactant is determined, and approximately 59 grams of sodium chloride (NaCl) can be produced.

1. Calculate the moles of each reactant:

Sodium: 23 g / 23 g/mol = 1 mol

Chlorine: 71 g / (2 * 35.5 g/mol) = 1 mol

2. Determine the limiting reactant:

In this case, both sodium and chlorine have the same number of moles (1 mol each). Therefore, neither is technically "limiting" the reaction. However, for calculating the theoretical yield (maximum amount of product), we need to consider that one reactant might be completely consumed before the other.

3. Calculate the grams of product based on the limiting reactant (assuming complete consumption of one reactant):

Since both reactants have the same amount, choosing either sodium or chlorine as the limiting reactant will give the same result. Let's assume sodium is completely consumed.

NaCl produced: 1 mol Na x (1 mol NaCl / 1 mol Na) x (23 g NaCl/mol + 35.5 g NaCl/mol) = 58.5 g NaCl

4. Round the answer to the nearest gram:

58.5 g rounded to the nearest gram is 59 g.

Therefore, if 23 g of sodium reacts completely with 71 g of chlorine, we can expect around 59 grams of sodium chloride (NaCl) to be produced.

If 23 g of sodium reacts with 71 g of chlorine, based on stoichiometry and the limiting reagent principle, 94 grams of sodium chloride would be the expected product, rounded to the nearest gram.

If 23 g of sodium reacts completely with 71 g of chlorine, the amount of product formed can be determined using stoichiometry based on the balanced chemical equation 2 Na(s) + Cl2(g) → 2 NaCl(s).

Based on the molar mass of the reactants and products, we can conclude that 45.98 amu of sodium will react with 70.90 amu of chlorine to produce 275.9 amu of sodium chloride. Converting these amounts to grams, we find that every 22.99 g of sodium reacts with 35.45 g of chlorine to produce 58.45 g of sodium chloride (NaCl).

Given the ratio of sodium to chlorine to sodium chloride is 1:1.545:2.54 in terms of grams, we can calculate that 23 g of sodium will react with 35.54 g (23 g × 1.545) of chlorine to produce a total of 58.54 g (23 g × 2.54) of sodium chloride. Since the question states that we have 71 g of chlorine, sodium is the limiting reagent, and therefore the maximum yield of NaCl will be based on the amount of sodium.

Thus, if 23 g of sodium reacts completely with 71 g of chlorine, 94 grams of sodium chloride (NaCl) would be the expected rounded product to the nearest gram.

Answer:

C. Nitrogen (N) and sulfur (S)

Explanation:

Hello,

In this case, we could differentiate the nonmetals from the metals by understanding they do not have or have very tiny values of properties such as bright, hardness, electric conductivity, heat conductivity and others. Moreover, they are allocated at the right of the periodic table. In such a way, since nitrogen tends to be a gas and sulfur a yellowish powder, they are classified as nonmetals whereas, lithium, barium, palladium, zinc, beryllium and magnesium are considered as metals as they have the aforementioned properties.

Best regards.

We can see here that the nonmetals among the options provided are:

C. Nitrogen (N) and Sulfur (S)

Nonmetals are a group of elements found on the right side of the periodic table. They are characterized by their properties, which are distinct from those of metals.

Nonmetals generally have properties such as being poor conductors of heat and electricity, having lower melting and boiling points, and being more brittle compared to metals. Nitrogen and sulfur are both nonmetals, whereas the other elements listed in the options (lithium, barium, palladium, zinc, beryllium, and magnesium) are metals or metal-like elements.

Learn more about nonmetals on https://brainly.com/question/16749127

#SPJ6

Final answer:

The balanced equation for the thermal dehydration of barium chloride is [tex]\[ BaCl_2 \cdot 2H_2O \xrightarrow[\text{}]{\text{Heat}} BaCl_2 + 2H_2O \uparrow \][/tex]

Explanation:

The thermal dehydration of barium chloride (BaCl₂) involves the removal of water molecules from its hydrated form. Barium chloride commonly exists as a dihydrate, BaCl₂·2H₂O. The balanced chemical equation for the thermal dehydration of barium chloride dihydrate is:

[tex]\[ BaCl_2 \cdot 2H_2O \xrightarrow[\text{}]{\text{Heat}} BaCl_2 + 2H_2O \uparrow \][/tex]

In this equation, the dihydrate on the left side loses two water molecules upon heating, producing anhydrous barium chloride (BaCl₂) and releasing water vapor. The upward arrow indicates the release of water in the form of steam or water vapor.

This process is a common example of thermal decomposition reactions, where a substance breaks down into simpler components upon exposure to heat. Understanding such reactions is crucial in various chemical and industrial processes, providing insights into the behavior of compounds under specific conditions.

Answer: water is the solvet, sugar is the solute and the the end result is a solution.

Explanation:

Binary Solution is a homogeneous mixture of two components called as solute and solvent.

Solute is the component which is present in smaller proportion and is solid for solid in liquid solution and solvent is the component which is present in larger proportion and is liquid for solid in liquid solution.

Concentrated solution is one in which there is more amount of solid.

Dilute solution is one in which there is more amount of liquid.

Thus as sugar is lesser in amount , it is the solute. Water is preset in larger amount, thus it is the solvent and the end result is the solution.

Answer:

to change the core

Explanation:

Answer:

6.0 × 10^23

Explanation:

Big brain mode

Answer:

B/ the second option

Explanation:

This is the answer on e2020

CAN YOU MAKE ME BRAINLIEST

Answer:

6 hope this helps