2Al + 6HCl → 2AlCl3 + 3H2
If 85.0 grams of HCl react, how many moles of H2 are produced?

Answer:

0.25 mole of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4Fe₃O₄ + O₂ —> 6Fe₂O₃

From the balanced equation above,

4 moles of Fe₃O₄ reacted with 1 mole of O₂.

Finally, we shall determine the number of mole of O₂ required to react with 1 mole of Fe₃O₄. This can be obtained as follow:

From the balanced equation above,

4 moles of Fe₃O₄ reacted with 1 mole of O₂.

Therefore, 1 mole of Fe₃O₄ will react with = (1/4) = 0.25 mole of O₂.

Thus, 0.25 mole of O₂ is required for the reaction.

Answer:

Mass of HCl leftover = 6.4 g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NaOH + HCl —> NaCl + H₂O

Next, we shall determine the masses of NaOH and HCl that reacted from the balanced equation. This can be obtained as follow:

Mass of NaOH = 23 + 16 + 1

= 40 g/mol

Mass of NaOH from the balanced equation = 1 × 40 = 40 g

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Mass of HCl from the balanced equation = 1 × 36.5 = 36.5 g

SUMMARY:

From the balanced equation above,

40 g of NaOH reacted with 36.5 g of HCl.

Next, we shall determine the mass of HCl required to react with 7 g of NaOH. This can be obtained as follow:

From the balanced equation above,

40 g of NaOH reacted with 36.5 g of HCl.

Therefore, 7 g of NaOH will react with = (7 × 36.5)/40 = 6.4 g of HCl.

Thus, 6.4 g of HCl is required for the reaction.

Finally, we shall determine the leftover mass of HCl. This can be obtained as follow:

Mass of HCl given = 12.8 g

Mass of HCl that reacted = 6.4 g

Mass of HCl leftover =?

Mass of HCl leftover = (Mass of HCl given) – (Mass of HCl that reacted)

Mass of HCl leftover = 12.8 – 6.4

Mass of HCl leftover = 6.4 g

Answer:

Titration is an analytical technique that is widely used in the food industry. It allows food manufacturers to determine the quantity of a reactant in a sample. For example, it can be used to discover the amount of salt or sugar in a product or the concentration of vitamin C or E, which has an effect on product colour.

Answer:

How is titration used in the industry?

Titration is an analytical technique that is widely used in the food industry. It allows food manufacturers to determine the quantity of a reactant in a sample. For example, it can be used to discover the amount of salt or sugar in a product or the concentration of vitamin C or E, which has an effect on product colour.

Answer:

60ug

Explanation:

If it has experienced two half lives, that means it has been halved twice. in that case, to undo it, just multiply it by two twice. 0.15ug * 2 = 0.30ug. 0.30ug * 2 = 0.60ug. Hope this helps.

Answer:

The Answer is - (B)

Explanation:

I Just took the test on edg

Answer:

I hope that helps. I did not round the answer at the end. Maybe you want the final answer in a specific decimal place.

Answer:

[tex]P_2=6.0atm[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the Boyle's law, since the temperature is constant and both volume and pressure change:

[tex]P_2V_2=P_1V_1[/tex]

Thus, we solve for the final pressure, P2, to obtain:

[tex]P_2=\frac{P_1V_1}{V_2}[/tex]

And plug in the given data to get:

[tex]P_2=\frac{2.0atm*12.0L}{4.0L}\\\\P_2=6.0atm[/tex]

Regards!

Answer:

Ka = 6.02x10⁻⁶

Explanation:

The equilibrium that takes place is:

We calculate [H⁺] from the pH:

Keep in mind that [H⁺]=[A⁻].

As for [HA], we know the acid is 0.66% dissociated, in other words:

We calculate [HA]:

Finally we calculate the Ka:

Answer: The heat absorbed by the water is 52.823 J.

Explanation:

Given: Mass of metal = 5.05 g

Specific heat of water = 4.184 [tex]J/g^{o}C[/tex]

Initial temperature = [tex]24.8^{o}C[/tex]

Final temperature = [tex]27.3^{o}C[/tex]

Formula used to calculate heat absorbed is as follows.

[tex]q = m \times C \times (T_{2} - T_{1})[/tex]

where,

q = heat

m = mass of substance

[tex]T_{1}[/tex] = initial temperature

[tex]T_{2}[/tex] = final temperature

Substitute the values into above formula as follows.

[tex]q = m \times C \times (T_{2} - T_{1})\\= 5.05 g \times 4.184 J/g^{o}C \times (27.3 - 24.8)^{o}C\\= 52.823 J[/tex]

Thus, we can conclude that heat absorbed by the water is 52.823 J.

Answer:

The correct answer is -

1. Latitude

2. Altitude

3. Bodies of water

4. Wind system

5. Amount of rainfall

Explanation:

Lattitude is the angle from the earth's equator towards the north or south to the poles. This is one of the important factors that lead to different seasons and define the temperature of the area.

Altitude is the elevation from the sea that also helps in affecting the seasons as the higher the altitude cooler the temperature. Water bodies like the sea affect the wind and climate as they are not able to absorb heat faster in comparison to the land.

Wind also affects the seasons and climate of the area as a large amount of warm air cools, it gives off  lots of rain over the land. Precipitation or rainfall determines and influences the climate.

Answer: If you want to make 10 moles of [tex]AlCl_{3}[/tex] then 30 moles of HCl are required in the given reaction.

Explanation:

The given reaction equation is as follows.

[tex]2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}[/tex]

Here, 6 moles of HCl reacts with 2 moles of [tex]AlCl_{3}[/tex]. This means that 1 mole of [tex]AlCl_{3}[/tex] is obtained as follows.

[tex]\frac{6}{2} HCl\\= 3 HCl[/tex]

Hence, 3 moles of HCl will give 1 mole of [tex]AlCl_3[/tex]. So, moles of HCl required to react with 10 moles of [tex]AlCl_{3}[/tex] is as follows.

[tex]1 mol AlCl_3 = 3 mol HCl\\10 mol AlCl_{3} = 3 \times 10 mol HCl\\= 30 mol HCl[/tex]

Thus, we can conclude that if you want to make 10 moles of [tex]AlCl_{3}[/tex] then 30 moles of HCl are required in the given reaction.

Answer:

Neutrons are all identical to each other, just as protons are. Atoms of a particular element must have the same number of protons but can have different numbers of neutrons.

Explanation:

Since the vast majority of an atom's mass is found its protons and neutrons, subtracting the number of protons (i.e. the atomic number) from the atomic mass will give you the calculated number of neutrons in the atom. In our example, this is: 14 (atomic mass) – 6 (number of protons) = 8 (number of neutrons).

Answer:

OH- is 1x 10^ + 3

Explanation:

- and - = +

Answer: The Ka expression for an aqueous solution of hypochlorous acid is [tex]K_{a} = \frac{[H_{3}O^{+}][OCl^{-}]}{[HClO]}[/tex].

Explanation:

The chemical formula of hypochlorous acid is HClO. So, when it is added to water (solvent) then its dissociation is as follows.

[tex]HClO + H_{2}O \rightarrow H_{3}O^{+} + Cl^{-}[/tex]

When we write the equilibrium constant for this reaction then it is called acid acid dissociated constant.

Hence, the expression for acid dissociation constant of this reaction is as follows.

[tex]K_{a} = \frac{[H_{3}O^{+}][OCl^{-}]}{[HClO]}[/tex]

Thus, we can conclude that the Ka expression for an aqueous solution of hypochlorous acid is [tex]K_{a} = \frac{[H_{3}O^{+}][OCl^{-}]}{[HClO]}[/tex].

Answer:

% yield = 200%

Explanation:

By using 145 gram of aluminium

Using 26.982 g/mol of Al, 101.96 g/mol of Aluminium oxide is produced

Using 145 grams of aluminium, the yield of Aluminium oxide  would have been = (101.96 g/mol/26.982 g/mol)/145 grams

= 547.9282485 gram

Now % yield = actual yield /theoretical yield * 100

Substituting the given values, we get -

% yield = (547.928/270) *100

% yield = 200%

Answer: The balanced equation is [tex]2SO_{2} + O_{2} + 2H_{2}O \rightarrow 2H_{2}SO_{4}[/tex].

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side is called a balanced chemical equation.

For example, [tex]SO_{2} + O_{2} + H_{2}O \rightarrow H_{2}SO_{4}[/tex]

Here, number of atoms present on reactant side are as follows.

Number of atoms present on product side are as follows.

To balance this equation, multiply [tex]SO_{2}[/tex] by 2 and [tex]H_{2}O[/tex] by 2 on reactant side. Also, multiply [tex]H_{2}SO_{4}[/tex] by 2 on product side.

Hence, the equation can be re-written as follows.

[tex]2SO_{2} + O_{2} + 2H_{2}O \rightarrow 2H_{2}SO_{4}[/tex]

Now, the number of atoms present on reactant side are as follows.

Number of atoms present on product side are as follows.

Since, the atoms present on both reactant and product side are equal. Hence, the equation is now balanced.

Thus, we can conclude that the balanced equation is [tex]2SO_{2} + O_{2} + 2H_{2}O \rightarrow 2H_{2}SO_{4}[/tex].

Answer:

a. molecules that gain or lose an atom to develop a bonding charge

Explanation:

Answer:

See explanation and image attached

Explanation:

2‑methylhept‑3‑yne can be made by the reaction of the compound shown in the image attached with soidamide in liquid ammonia and 1‑bromopropane.

The first step in the reaction is the deprotonation of the alkyne starting material by the amide ion.

This step is followed a nucleophilic substitution reaction of the alkyne with 1‑bromopropane to yield the 2‑methylhept‑3‑yne product.

Answer:

Explanation:

The chemical reaction of the well-balanced equation is as follows:

[tex]\mathbf{2NiCl_{2(aq)}+2Na_2S_{(aq)} \to 4NaCl_{(aq)}+2NiS_{s}}[/tex]

The aforementioned reaction's ionic equation is:

[tex]\mathbf{2Ni^{2+}_{(aq)}+4Cl^-_{(aq)}+ 4Na^+_{(aq)}+2S^{2-}_{(aq)} \to 4Na^+_{(aq)}+4Cl^-_{(aq)}+2NiS_{(s)}}[/tex]

From above; Na⁺ and Cl⁻ appear on both sides;

Hence, the net ionic equation is:

[tex]\mathbf{2Ni^{2+}_{(aq)}+2S^{2-}_{(aq)} \to 2NiS_{(s)}}[/tex]

Explanation:

Magnesium reacts with dilute hydrochloric acid in a conical flask which is connected to an inverted measuring cylinder in a trough of water. The volume of hydrogen gas produced is measured over a few minutes, and the results are used to plot a graph

This is intended as a class practical. It is best if the students work in pairs because setting up and starting the experiment requires more than one pair of hands. One student can add the magnesium ribbon to the acid and stopper the flask, while the other starts the stopclock. During the experiment, one student can take the readings while the other records them. The experiment itself takes only a few minutes. But allow at least 30 minutes to give students time to set up, take readings and draw graph.

please mark as brainliest

Answer:

Chloroform.

Explanation:

Given,

Solvent requires 1g of compound per 100 mL

For water,

= 1g/47ml

= 2.1

For Chloroform,

= 1 g/8.1 mL

= 12.345679

For Diethyl ether,

= 1 g/370 mL

= 0.27

For Benzene,

=  1 g/86 mL

= 1.2

Partition coefficients:

Water = -

chloroform = 5.9

Diethyl = .13

Benzene  = .57

The solvent chloroform would be chosen for drawing out the compound out of an aqueous solution as it has the maximum solubility.

Answer:

Sodium arachidate; Sodium palmitate and Sodium palmitate

Explanation:

Triglycerides are esters of fatty acids with glycerol. In triglycerides, three fatty acid molecules are linked by ester bonds to each of the three carbon atoms in a glycerol molecule. The fatty acids may be same or different fatty acid molecules. Hydrolysis of triglycerides yields the three fatty acid molecules and glycerol.

Saponification is the process by which a base is used to catalyst the hydrolysis of the ester bonds in glycerides. The products of this base-catalyzed hydrolysis of triglycerides are the metallic salts of the three fatty acids and glycerol. The salts of the fatty acids are known as soaps.

For a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic acid attached to the three backbone carbons glycerol, the saponification of the triglyceride with NaOH will yield the sodium salts or soaps of the three fatty acids as well as glycerol.

Arachidic acid will react with NaOH to yield sodium arachidate.

The two palmitic acid molecules will each react with NaOH to yield sodium palmitate.

Answer:

See explanation.

Explanation:

Hello there!

In this case, for this titration problem, we first need to set up the undergoing chemical equation between barium chloride and hydrochloric acid:

[tex]2HCl+Ba(OH)_2\rightarrow BaCl_2+2H_2O[/tex]

Which occurs in a 2:1 mole ratio of acid to base and thus, we can write the following:

[tex]2n_{base}=n_{acid}[/tex]

Which can be written in molarities and volumes:

[tex]2M_{base}V_{base}=M_{acid}V_{acid}[/tex]

And we solve for the molarity of the acid:

[tex]M_{acid}=\frac{2M_{base}V_{base}}{V_{acid}}[/tex]

Unfortunately, the molarity of the base was not given:

"The molarity of an aqueous solution of hydrochloric acid, , is determined by titration with a M barium hydroxide, , solution. If 31.2 mL of barium hydroxide are required to neutralize 15.4 mL of the acid, what is the molarity of the hydrochloric acid solution?"

Yet we can assume an arbitrary value, say 1.0 M in order to show you the solution so that you can modify it according to your given value:

[tex]M_{acid}=\frac{2(1.0M)(31.2mL)}{(15.4mL)}=4.05M[/tex]

So you just need to modify the (1.0 M) by the actually given molarity of the acid.

Regards!

Answer:

c = π/T*R

c = 0.199M

Explanation:

Is measured to be 5.1 atm at an absolute temperature of 312. K. Write an equation that will let you calculate the molarity c of this solution. Your equation should contain only symbols. Be sure you define each symbol other than R.

The osmotic pressure of a solution, π, is defined as:

π = c*T*R

As we know osmotic pressure π = 5.1atm, T = 312K and R = 0.082atmL/molK

we can replace in the equation as follows:

π= c*T*R

c = 5.1atm/312K*0.082atmL/molK

Answer: For the given reaction [tex]Pb(NO_{3})_{2} + AlBr_{3}[/tex] the products are [tex]PBr_{2} + Al(NO_{3})_{3}[/tex].

Explanation:

A reaction in which two substances chemically combine together and results in the formation of a new substance is called a chemical reaction.

For example, [tex]3Pb(NO_{3})_{2} + 2AlBr_{3} \rightarrow 2Al(NO_{3})_{3} + 3PbBr_{2}[/tex]

Products are the species that are written on the right side of an arrow in a chemical reaction equation.

Hence, we can conclude that for the given reaction [tex]Pb(NO_{3})_{2} + AlBr_{3}[/tex] the products are [tex]PBr_{2} + Al(NO_{3})_{3}[/tex].