A Virginia class nuclear submarine has an internal volume of 7.9 million
liters at a pressure of 1.0 atm. If a crewman were to open one of the
hatches to the outside ocean while it was underwater (pressure = 25 atm),
what be would the new volume of the air inside the submarine?

Answer:

D

Explanation:

Valency is the combining power of an element

The molar mass of the unknown compound is 35.38 g/mol.

PV = nRT

First, we need to convert the volume from mL to L:

291 mL = 0.291 L

Next, we can solve for the number of moles of the unknown compound:

n = PV/RT = (2.93 atm)(0.291 L)/(0.08206 L atm/mol K)(298 K) = 0.0145 mol

molar mass = mass/number of moles = 0.513 g/0.0145 mol = 35.38 g/mol

Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance. It is usually expressed in units of grams per mole (g/mol). A mole is a unit of measurement used to express the number of atoms or molecules in a substance. One mole of any substance contains Avogadro's number of particles, which is approximately 6.022 x \(10^{23\).

Molar mass is important in chemical calculations, as it allows chemists to convert between mass and moles of a substance. This is useful in determining the amount of reactants needed in a chemical reaction, or the amount of product produced. Additionally, molar mass is used in the calculation of various other important properties of a substance, such as density, specific heat, and concentration.

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Answer:

Chemical reaction, not physical

Explanation:

first you need find number of mole

number of mole =given mass/ molar mass. 6.90g/16g/mol =0.431mol . mass =molar mass × number of mole KNO3 = 101g/mol × 0.431 mol = 43.5g .

Here is an example

Explanation:

From n = 5 , the electron can drop to n =

4, 3, 2, 1. Thus, there will be 4

possibilities

Now, the electron that dropped to n = 4 can drop to n =3,2,1 This will give 3 possibilities

The electron at n = 3 can drop to n =2,1 This gives 2 possibilities

Finally, the electron at n = 2 can drop to n = 1 . This gives 1 possibility

Thus in total the number of ways in which the electron can de-excite is :

4 + 3 + 2 + 1 = 10

10 different wavelengths of light can be emitted as the electron loses energy

from n = 5 level

Answer:

Tin and Copper

Explanation:

The two atoms that could potentially form alloys are the tin and copper atoms.

Alloys are mixtures formed between two or more metals.

The goal of forming alloys is to take properties of one metal and add to that of another metal so as to enhance both materials.

Alloys are only formed between metals.

The only metal pair given from the choices is that of tin and copper.

Therefore, the two atoms that could potentially form alloys is tin and copper.

The IR spectrum of Acetylferrocene will differ from that of Ferrocene because of the carbonyl stretching and bending modes that appear in the Acetylferrocene.

The IR spectrum of Acetylferrocene will differ from that of Ferrocene due to the carbonyl stretching and bending modes that appear in the Acetylferrocene.

Ferrocene and Acetylferrocene have similar IR spectra since they both have the Fe-Cp stretching and bending modes. The acetyl group of Acetylferrocene is reflected by an intense band in the 1700-1750 cm-1 range, which is due to carbonyl stretching.

                                     In Acetylferrocene, the IR spectra are dominated by the presence of the acetyl group's vibration, resulting in a change in the frequency of stretching vibration from 200 to 220 cm−1. Another change in the IR spectra of Acetylferrocene is the presence of two bands due to C-O stretching at 1230-1260 cm-1 in addition to the appearance of a strong band due to C-H bending vibrations in the 1410-1450 cm-1 region.

                           Ferrocene does not have a carbonyl group, which is why it will not display the carbonyl stretching and bending vibrations in the IR spectra. This is the most significant difference between the two IR spectra. So, we can conclude that the IR spectrum of Acetylferrocene will differ from that of Ferrocene due to the presence of the carbonyl group.

Therefore,  The IR spectrum of Acetylferrocene will differ from that of Ferrocene because of the carbonyl stretching and bending modes that appear in the Acetylferrocene.

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Decreasing the TBCl concentration will not affect the rate constant in this experiment. The rate constant is determined by the specific reaction and temperature conditions and is independent of the reactant concentrations.

The rate constant (k) is a proportionality constant that relates the rate of a reaction to the concentrations of the reactants. However, the rate constant itself is not affected by the concentrations of the reactants. It is determined by the specific reaction and temperature conditions.

The rate of a chemical reaction can be expressed using the rate equation, which typically includes the concentration terms for the reactants raised to certain powers.

These powers, known as reaction orders, can be determined experimentally. However, the rate constant is a separate factor in the rate equation and is not dependent on the reactant concentrations.

By decreasing the TBCl concentration, the rate of the reaction may be affected, as the rate is directly proportional to the reactant concentrations.

However, the rate constant itself remains unchanged. The rate constant is influenced by factors such as temperature, presence of catalysts, and the nature of the reacting species, but not by the concentrations of the reactants.

Therefore, decreasing the TBCl concentration will not affect the rate constant in this experiment.

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The mole fraction of hydrogen is 0.55.

The volume of oxygen = 5 litres

The pressure of oxygen = 2 atm

The volume of hydrogen = 3 litres

The pressure of hydrogen = 4 atm

The universal gas constant is R.

R = 8.31 J mol—¹ K—¹

The ideal gas equation is,

PV = nR T

The number of moles of oxygen is,

PV = nRT

\(n = \frac{ PV }{RT }\)

\(n = \frac{2 \times 5}{8.31 \times 273} \)

n = 0.004 moles

The number of moles of oxygen is 0.004 moles.

The number of moles of hydrogen is,

PV = nRT

\(n = \frac{ PV }{RT }\)

\(n = \frac{4 \times 3}{8.31 \times 273} \)

n = 0.005 moles

The number of moles of hydrogen is 0.005 moles.

The mole fraction of hydrogen is,

\(Mole \: fraction = \frac{ Moles \: of \: solute }{Total \: number \: of \: moles \: of \: solute \: and \: solvent}\)

\(Mole \: fraction = \frac{0.005}{0.009}\)

= 0.55

Therefore, the mole fraction of hydrogen is 0.55.

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Answer:

Iron is oxidized to form rust.

Explanation:

Consider the reaction; 4Fe + 302------>2Fe2O3, we can see that iron is being oxidized to iron III oxide. Rust is the common name of iron III oxide.

Rusting is an electrochemical process, iron rusts when it comes into contact with air and water because electrochemical cells are set up at the surface of contact.

Iron usually functions as the anode in the electrochemical process. This process leads to the formation of iron III oxide. Rust is soft and breaks off easily thereby exposing the metal below the surface to further rusting.

Answer:

Iron is oxidized to form rust.

Explanation:

edge

Coefficient 2 should be placed in the equation before the nitric acid in the reactant so that the chemical reaction becomes balanced.

The given chemical reaction is,

Zn + HNO3 --------> Zn (No3)2 + H2

Here, the elements Hydrogen, Nitrogen, and Oxygen are unbalanced.

These elements are altogether found in HNO₃ on the reactant side and are found as nitrate and hydrogen gas on the product side.

To balance the equation, the coefficient 2 is to be added in the reactant side so that 2 moles of Hydrogen, Nitrogen, and Oxygen would be balanced with 2 moles of Nitrate and two moles of Nitrogen.

On adding 2 on reactant side, we get

Zn + 2HNO3 -----------> Zn(NO3)₂ + H₂

This gives us a balanced equation and thus the equation could be balanced this way.

Therefore, coefficient 2 is to be added to make the equation to be balanced.

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The chemist should use 30 liters of the 20% solution, 20 liters of the 30% solution, and 40 liters of the 60% solution to obtain a mixture of 90 liters containing 40% acid.

Let's denote the number of liters of the 20% solution as x, the number of liters of the 30% solution as y, and the number of liters of the 60% solution as z.

According to the problem, we have the following information:

The total volume of the mixture is 90 liters: x + y + z = 90.

The desired percentage of acid solution in the mixture is 40%: (0.20x + 0.30y + 0.60z) / 90 = 0.40.

The chemist wants to use 2 times as much of the 60% solution as the 30% solution: z = 2y.

Now, we can solve these equations simultaneously to find the values of x, y, and z.

From equation 3, we have z = 2y.

Substituting this into equation 1, we get:

x + y + 2y = 90

x + 3y = 90 (equation 4)

From equation 2, we have:

(0.20x + 0.30y + 0.60z) / 90 = 0.40

Multiplying both sides by 90, we get:

0.20x + 0.30y + 0.60z = 36

0.20x + 0.30y + 0.60(2y) = 36 (since z = 2y)

0.20x + 0.30y + 1.20y = 36

0.20x + 1.50y = 36 (equation 5)

Now, we can solve equations 4 and 5 simultaneously to find the values of x and y.

Multiplying equation 4 by 0.20, we get:

0.20(x + 3y) = 0.20(90)

0.20x + 0.60y = 18

Subtracting this from equation 5, we eliminate x and solve for y:

0.20x + 1.50y - (0.20x + 0.60y) = 36 - 18

0.90y = 18

y = 18 / 0.90

y = 20

Substituting this value of y back into equation 4, we find:

x + 3y = 90

x + 3(20) = 90

x + 60 = 90

x = 90 - 60

x = 30

Since z = 2y, we have:

z = 2(20)

z = 40

Therefore, the chemist should use 30 liters of the 20% solution, 20 liters of the 30% solution, and 40 liters of the 60% solution to obtain a mixture of 90 liters containing 40% acid.

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The specific heat of calcium carbonate for 5.04g sample that absorbs 40 J of heat and rises in temperature from
24.5 °C to 34.1 °C is  0.83 J/(g·°C)

To determine the specific heat of calcium carbonate (CaCO3), we can use the following formula:
Specific Heat (c) = Heat absorbed (Q) / (mass (m) × change in temperature (ΔT))
Given data:
Heat absorbed (Q) = 40 J
Mass (m) = 5.04 g
Initial temperature = 24.5 °C
Final temperature = 34.1 °C
First, we need to find the change in temperature (ΔT):
ΔT = Final temperature - Initial temperature = \(34.1^\circ C- 24.5^\circ C = 9.6 ^\circ C\)
Now, we can calculate the specific heat (c):
c = \(40 J / (5.04 g * 9.6^\circ C) =0.83J/ (g^\circ C)\)

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"The most reactive metals are typically located in the far left area of the periodic table, also known as the alkali metals." These metals include elements like lithium, sodium, and potassium.

Alkali metals are highly reactive because they have only one electron in their outermost energy level, which they readily lose to form positive ions. Key characteristics of alkali metals include:

1. Reactivity: Alkali metals are the most reactive metals. They readily lose their outermost electron to form a +1 ion, making them highly reactive with other elements.

2. Softness: Alkali metals have low hardness and can be easily cut with a knife.

3. Low density: They have low densities compared to other metals.

4. Low melting and boiling points: Alkali metals have relatively low melting and boiling points.

5. Good conductors of heat and electricity: They are efficient conductors of heat and electricity.

6. Reactivity with water: Alkali metals react vigorously with water, producing hydrogen gas and hydroxide ions.

7. Oxidation: Alkali metals readily react with oxygen in the air, forming oxides or peroxides.

8. Flame coloration: Alkali metals, when heated, produce distinct colors in flames. For example, sodium imparts a yellow color, and potassium gives a lilac color.

Alkali metals are important in various applications, such as batteries, alloys, and certain chemical reactions. However, their high reactivity makes them challenging to handle safely, requiring special precautions due to their tendency to react explosively with moisture or air.

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The name of the compound made of magnesium and fluorine is magnesium fluoride and its formula is MgF2. Magnesium fluoride is a white crystalline solid that has a bitter taste and is insoluble in water. Magnesium fluoride is a stable ionic compound with a high melting point.

This compound is the primary source of magnesium and fluoride ions in many industrial and chemical processes. Magnesium fluoride is produced by the reaction between magnesium oxide and hydrogen fluoride.

MgO + 2HF → MgF2 + H2O

In the laboratory, magnesium fluoride is produced by dissolving magnesium hydroxide in hydrofluoric acid.

Mg(OH)2 + 2HF → MgF2 + 2H2O

Magnesium fluoride is used in the production of ceramics and glass as a flux and to increase the resistance of the glass to attack by acids. Magnesium fluoride is also used as an optical coating on lenses and other optical components as it has a low refractive index.

Magnesium fluoride is a white crystalline solid with a bitter taste that is insoluble in water. It is the primary source of magnesium and fluoride ions in many industrial and chemical processes. The formula of magnesium fluoride is MgF2. It is used in the production of ceramics and glass as a flux and to increase the resistance of the glass to attack by acids. Magnesium fluoride is also used as an optical coating on lenses and other optical components as it has a low refractive index.

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Explanation:

a. qualitative

b. quantitative

c. quantitative

d. quantitative

Hope this helps you.

Water-soluble hydroxides of metals from Groups 1A and 2A are considered strong bases because they are ionic compounds that dissociate completely in aqueous solution.

The hydroxide ion (OH-) is a strong base because it can accept a proton (H+) from water to form a hydroxide ion (OH-) and a hydronium ion (H3O+). When the hydroxide ions from the metal hydroxides dissolve in water, they combine with the hydronium ions to form water molecules and the corresponding metal ions. The resulting solution is alkaline because of the excess of hydroxide ions.

These strong bases have various uses in industry and laboratory settings. For example, sodium hydroxide (NaOH) is used in the manufacture of soap, paper, and aluminum, while calcium hydroxide (Ca(OH)2) is used in water treatment, agriculture, and construction. When handling these strong bases because they can cause burns and other injuries. It is also important to dispose of them properly to prevent environmental damage.

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We can begin by completing the appropriate algebraic operations to isolate x on one side of the equation before trying to identify it.

20 = 80(.5)^8.32/x

Put 80 on both sides:

20/80 = (.5)^8.32/x

Make the left side simpler:

1/4 = (.5)^8.32/x

Take both sides' reciprocal:

4 = x/(.5)^8.32

Make the right side simpler:

4 = x/0.0027067

Divide the two sides by 0.0027067:

x = 0.0027067 * 4

x = 0.0108268

Hence, x is almost equal to 0.0108268.

Your response is in scientific notation since 5.94 is at least 1 but less than 10. (3.4 10 3) + (5.6 10 4) = 5.94 10 4 is the solution.

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Answer:

0.000471698

Explanation:

divide 0.125 by 265.

Answer:

light

Explanation:

Have a nice day

Answer:

light

Explanation:

Answer:

0.8338395752301793

Explanation:

I used an online converter (grams to moles for FeCl2)

The process that was used in this lab to create magnesium oxide by using oxidation type of reaction and it relates to conservation of matter

The total mass of reactants and products in a chemical reaction must equal one another, according to the law of conservation of mass.

1. The method used in the laboratory to produce MgO

According to the following equation, Mg is weighed in a crucible and heated in air that strongly reacts with O₂ to produce MgO:

1/2O₂ + Mg + = MgO

and the growth in the mass of magnesium in the crucible supports this.

2. The kind of chemical reaction that occurred:

Since the magnesium gained O and was oxidized to MgO, it is thought to be an oxidation reaction.

3. The following explains why the product was heavier than the reactant:

The product had a higher mass due to the gain of O atom over the reactant Mg because the reactant (Mg) bonded to O and formed MgO.

4. In relation to the conversation of matter:

According to the law of conversation of matter, matter can be changed from one form to another but cannot be destroyed or created from nothing.

Here, Mg and O2 changed into MgO, fixing the presence of matter and the conversation of mass.

If we have 1.0 mole of magnesium, it will react with 0.5 mole of oxygen to form 1.0 mole of magnesium oxide.

The Mg mass is = n x molar mass of Mg

The Mg mass is= (1.0 mole) (24.30 g/mol)

The Mg mass is = 24.3 g.

The mass of O₂ is =  n x molar mass of O₂

The mass of O₂ is= (0.5 mole) (32.00 g/mole)

The mass of O₂ is = 16.0 g.

The MgO mass is = n x molar mass of MgO

The MgO mass is  = (1.0 mole) (40.30 g/mol)

The MgO mass is = 40.3 g/mol.

The reactants mass is = The mass of Mg + The mass of O₂

The reactants mass is = 24.3 g + 16.0 g = 40.3 g.

The  products mass is = The mass of MgO = 40.3 g.

∴ The mass of the reactants is equal to the mass of the products.

Therefore, there is a conservation of matter.

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Complete question is " Describe the process that was used in this lab to create magnesium oxide, specifically identifying the type of chemical reaction. Explain why the product had a higher mass than the reactant, and how this relates to conservation of matter".

If a solution contains both a strong electrolyte and a nonelectrolyte and is tested with a conductivity apparatus, the observation would be a bright light in the light bulb. Option(b)

A strong electrolyte dissociates completely into ions in solution, leading to the presence of free-moving charged particles that can conduct electricity. This results in a strong electrical current and a bright light in the conductivity apparatus.

On the other hand, a nonelectrolyte does not dissociate into ions and does not contribute to conductivity. However, in the presence of a strong electrolyte, the conductivity is dominated by the electrolyte, leading to a bright light in the bulb. Therefore, the correct answer is (b) bright light.

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Answer:

Single replacement

Explanation:

The percent composition by mass of oxygen in magnesium sulfate (MgSO₄)  is 53.33%.

The percent composition of mass of oxygen in magnesium sulfate can be calculate as follows

we know that Molar mass of magnesium sulfate = 120 g/mol

Magnesium sulfate has four oxygen atoms, so multiply the molar mass of oxygen by four. So 16 × 4 = 64 g/mol.

To calculate the percent oxygen composition,

% oxygen composition = mass molar oxygen / mass molar

% oxygen composition = 64/120 × 100

% oxygen composition = 53.33 %

Therefore, the percent composition by mass of oxygen in magnesium sulfate (MgSO₄)  is 53.33%.

The mass fraction of every element found in a compound is referred to as the percentage composition of the solution.

Three techniques can be used to determine the percentage composition of matter:

The first method is good if both ingredients are solid and can be measured in grams.

A second approach can be used when both components are liquids.

A third approach is used when one substance is solid and the other is liquid. Divide the two quantities and multiply by 100 to calculate the composition ratio. By convention, both quantities must have the same units.

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