Calcium chloride (aq) reacts with sodium carbonate (aq) to from solid calcium carbonate and aqueous sodium chloride. Determine the volume of a 2.00 M Calcium chloride solution would be needed to exactly react with 0.0650 L of 1.50 M Na2CO3. (Use BCA!)

Answer:

E. activated complex

Explanation:

Answer:

see explanation

Explanation:

Using the periodic table, look at the top of each column => These are group numbers ... Typically (for american tables) the numbers are divided into A-Groups and B-Groups... For this post, you need to focus on the A-Groups, or 'Main Group Elements'... Now, the numbers also represent the number of valence (outer shell) electrons in the elements listed below that number. For example, under Group IA, all elements ( H, Li, Na, K, Rb, Cs & F) all have one (1) outer shell electron. All elements under IIA have two outer shell electrons, IIIA, 3 outer shell electrons and so on. The exception is Helium (He) which has only 2 outer shell electrons and is typically listed under Group VIIIA.

So ...

X· => H, Li & Na

X: => He(noble gas exception), Be & Mg

·X: => B & Al

:X: => C & Si

X(5 dots) => N  

X(6 dots) => O

X(7 dots) => F & I

X(8 dots) => Ne

Final answer:

It is true that plasma spray-coating processes are used to protect materials from environments which could cause degradation due to factors like wear, corrosion, or thermal failure. Corrosion, a REDOX process, can be combatted with protective coatings, including more nobel metals or cathodic protection via sacrificial anodes.

Explanation:

True, plasma spray-coating processes are indeed utilized to provide surface protection for materials that are exposed to hostile environments. These environments may induce degradation through wear, corrosion, or thermal failure. Corrosion, specifically, is a REDOX (reduction-oxidation) process where metals deteriorate through oxidation. This fact is evident in how iron rusts or how copper develops a patina when exposed to air. Protective coatings are a common method to prevent corrosion, as they can consist of a second metal that is more difficult to oxidize, or contain more easily oxidized metals, providing cathodic protection. For instance, the application of a thin layer of zinc can protect galvanized steel and sacrificial electrodes can be attached to protect metals such as iron.

Answer : The correct option is, (c) use of a mobile and a stationary phase.

Explanation :

Chromatography : It is a separation process or technique of a mixture in which a mixture is distributed between the two phases at different rates, one of which is stationary phase and another is mobile phase.

Mobile phase : The mixture is dissolved in a solution is known as mobile phase.

Stationary phase : It is an adsorbent medium and It is a solid, liquid or gel that remains immovable when a liquid or a gas moves over the surface of adsorbent. It remains stationary.

Hence, a characteristic feature of any form of chromatography is the use of a mobile and a stationary phase.

Answer:

The answer to your question is 1) 0.037 M  2) 0.32 M  3)  0.096 M

Explanation:

a) 125 ml of 0.251 M HCl

-Calculate the moles of HCl

Molarity = moles/volume

-Solve for moles

moles = Molarity x volume

-Substitution

moles = 0.251 x 0.125

           = 0.0314

-Calculate the new molarity

Molarity = 0.0314/ (0.125 + 0.250)

-Simplification

Molarity = 0.014/0.375

-Result

Molarity = 0.037 M

2.-

445 ml of 0.499 M of H₂SO₄

-Calculate the number of moles

moles = 0.499 x 0.445

moles = 0.222

-Calculate the new molarity

Molarity = 0.222/(0.445 + 0.25)

Molarity = 0.222/0.695

Molarity = 0.32

3)

5.25 l of HCO₃ 0.101 M

Calculate the number of moles

moles = 0.101 x 5.25

moles = 0.53

-Calculate the Molarity

Molarity = 0.53 / (0.25 + 5.25)

Molarity = 0.53 / 5.5

Molarity = 0.096

The new molarity of each question is: (a): 0.0835 M, (b): 0.3194 M, and (c) 0.0964 M.

The new molarities after adding 250 mL of water to each solution are: a) 0.0835 M for the first solution (HCI) b) 0.3194 M for the second solution (H2SO4) c) 0.0964 M for the third solution (HCO3).

To calculate the new molarity after dilution, we can use the dilution equation:

[tex]\[ C_1V_1 = C_2V_2 \][/tex]

where [tex]\( C_1 \)[/tex] is the initial concentration, [tex]\( V_1 \)[/tex] is the initial volume, [tex]\( C_2 \)[/tex] is the final concentration, and [tex]\( V_2 \)[/tex] is the final volume.

Let's solve for each case:

a) For the first solution:

- Initial concentration [tex](\( C_1 \))[/tex] = 0.251 M

- Initial volume [tex](\( V_1 \))[/tex] = 125 mL

- Final volume [tex](\( V_2 \))[/tex] = 125 mL + 250 mL = 375 mL

We need to find the final concentration [tex](\( C_2 \))[/tex]. Using the dilution equation:

[tex]\[ C_1V_1 = C_2V_2 \][/tex]

[tex]\[ 0.251 \text{ M} \times 125 \text{ mL} = C_2 \times 375 \text{ mL} \][/tex]

[tex]\[ C_2 = \frac{0.251 \text{ M} \times 125 \text{ mL}}{375 \text{ mL}} \][/tex]

[tex]\[ C_2 = \frac{31.375 \text{ mL} \cdot \text{M}}{375 \text{ mL}} \][/tex]

[tex]\[ C_2 = 0.0835 \text{ M} \][/tex]

So, the new molarity for solution A is 0.0835 M.

b) For the second solution:

- Initial concentration [tex](\( C_1 \))[/tex] = 0.499 M

- Initial volume [tex](\( V_1 \))[/tex] = 445 mL

- Final volume [tex](\( V_2 \))[/tex] = 445 mL + 250 mL = 695 mL

Using the dilution equation:

[tex]\[ C_1V_1 = C_2V_2 \][/tex]

[tex]\[ 0.499 \text{ M} \times 445 \text{ mL} = C_2 \times 695 \text{ mL} \][/tex]

[tex]\[ C_2 = \frac{0.499 \text{ M} \times 445 \text{ mL}}{695 \text{ mL}} \][/tex]

[tex]\[ C_2 = \frac{222.005 \text{ mL} \cdot \text{M}}{695 \text{ mL}} \][/tex]

[tex]\[ C_2 = 0.3194 \text{ M} \][/tex]

So, the new molarity for solution B is approximately 0.3194 M.

c) For the third solution:

- Initial concentration [tex](\( C_1 \))[/tex] = 0.101 M

- Initial volume [tex](\( V_1 \))[/tex] = 5.25 L = 5250 mL

- Final volume [tex](\( V_2 \))[/tex] = 5250 mL + 250 mL = 5500 mL

Using the dilution equation:

[tex]\[ C_1V_1 = C_2V_2 \][/tex]

[tex]\[ 0.101 \text{ M} \times 5250 \text{ mL} = C_2 \times 5500 \text{ mL} \][/tex]

[tex]\[ C_2 = \frac{0.101 \text{ M} \times 5250 \text{ mL}}{5500 \text{ mL}} \][/tex]

[tex]\[ C_2 = \frac{530.25 \text{ mL} \cdot \text{M}}{5500 \text{ mL}} \][/tex]

[tex]\[ C_2 = 0.0964 \text{ M} \][/tex]

So, the new molarity for solution C is approximately 0.0964 M.

Complete Question

Identify the orbitals that overlap to form the C−Cl bonds in CH2Cl2.

a. carbon sp3 hybrid orbital with a singly occupied chlorine 3s orbital

b. carbon sp2 hybrid orbital with a singly occupied chlorine 3p orbital

c. carbon sp3 hybrid orbital with chlorine sp3  hybrid orbital

d. carbon sp3 hybrid orbital with a singly occupied chlorine 3p orbital

e. a singly occupied carbon 2p orbital with chlorine sp3 hybrid orbital

Answer:

Correct Option is D

Explanation:

         The molecular formula of the compound given is

                               [tex]CH_2 Cl_2[/tex]

The structural formula for this given compound is shown on the first uploaded image

         Looking at the structural formula we see that all the bonds are single bonds  which shows that carbon is sp3 hybridzied which means that one 2s orbital of carbon has mixed with 3 2p orbital to for a form a four hybrid orbital as shown on the second uploaded image

  For the clorine the outer shell is containing two 3s orbital which are completely filled and a 6 3p orbital which requires an electron to complete it as shown on the uploaed image

      Hence the bond between the carbon and the clorine is between a

      sp3 hybridzied orbital and a 3p orbital

Note: each orbital contains a single electron

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Answer:

la neta no se me llevo los puntos

Explanation:

no se ingles

The temperature at which the volume of the gas increases from 49 mL to 74 mL, keeping the pressure constant, is approximately 149.43 0C, calculated using Charles's Law.

The question involves the application of Charles's Law, which relates the volume of a gas to its temperature when the pressure and the amount of gas are held constant. To find the temperature at which the volume of gas increases from 49 mL to 74 mL, you need to set up a ratio using Charles's Law, which is V1/T1 = V2/T2, where T is in Kelvin. Make sure to convert Celsius to Kelvin by adding 273.15.

First, we'll convert the initial temperature from Celsius to Kelvin: T1 = 7.0 + 273.15 = 280.15 K. Next, we'll use the volumes provided (V1 = 49 mL and V2 = 74 mL) to find the final temperature, T2.

By re-arranging Charles's Law, we get T2 = (V2 × T1) / V1. Plugging the values in, T2 = (74 mL × 280.15 K) / 49 mL = 422.58 K. To present this temperature in Celsius, subtract 273.15 from it: T2 = 422.58 K - 273.15 = 149.43 °C.

Therefore, the temperature at which the volume of gas is 74 mL is approximately 149.43 °C.

Answer:

Explanation:By the time, that the balloon got too high in the sky, the pressure inside the balloon will soon overcome the pressure outside, and the balloon's elasticity is not too strong to hold the air inside, that the Helium gas inside will successfully push out the walls of the balloon and so it bursts!

Amount of CO₂ emission per day is 11,356.23 g.

Explanation:

Joe travelling distance per day = 60 miles

Carbon dioxide emission per day = 20 mpg

Now we have to find the amount of carbon dioxide emitted per day by dividing the distance by the emission per day given in gallons.

Amount of Carbon dioxide emission = [tex]$\frac{distance}{emission amount}[/tex]

Amount of CO₂ emission in gallons  = [tex]$\frac{ 60 miles}{20 mpg}[/tex]  

                                                            = 3 gallons

Now we have to convert the gallons to grams as,

1 gallon = 3,785.41 g

3 gallons = 3 × 3785.41 g = 11,356.23 g

So the emission of CO₂ per day is 11,356.23 g.

Answer:

0.5 atm is equal to 380mmHg.

Explanation:

For every 1 atm, it is equal to 760mmHg.

Therefore, 0.5 atm is 760/2, which is 380mmHg.

Answer:

D) One half of the carbon atoms of newly synthesized acetyl CoA.

Explanation:

It will be radioactively labeled because Malonyl CoA which contains 3 Carbon molecule is synthesized from Acetyl CoA which has 2 Carbon molecule.

This happens with the addition of ‘CO2’ with the help of the enzyme called acetyl CoA carboxylase.

Final answer:

The compounds that will become radioactively labeled after inhaling 14C-labeled carbon dioxide are the carboxyl atoms of newly synthesized fatty acids, as this is the only portion derived directly from CO₂. Thus, option C is correct.

Explanation:

To determine which compounds will become radioactively labeled after industrial workers breathe air containing 14C-labeled carbon dioxide, we must understand the metabolic pathways that involve the incorporation of carbon from CO₂. During the synthesis of fatty acids, acetyl CoA serves as the two-carbon donor in the form of its activated methyl group, while malonyl CoA provides a two-carbon unit that loses one carbon as CO₂ during the elongation cycle. Though both are involved, individual carbon atoms from CO₂ do not directly become part of the fatty acid chain.

Given that the carboxyl group of the fatty acids does not originate from acetyl CoA or malonyl CoA, the correct choice would be C) The carboxyl atom of newly synthesized fatty acids. This is because the carboxyl group is the only portion derived from the original CO₂ breathed by the workers.

Answer:

The answer to your question is below

Explanation:

Ionic bond is a kind of bond in which a metal attaches to a nonmetal. Also we know that a molecule has ionic bonding if the electronegativity is higher than 1.7.

                    Kind of elements         Difference of electronegativity     Bond

a) MgO         Metal - Nonmetal                3.44 -  1.31 = 2.13                     Ionic

b) SrCl₂        Metal -Nonmetal                  3.16 - 0.95 = 2.21                    Ionic

c) O₃             Nonmetal- Nonmetal          3.44 - 3.44 = 0                      Covalent

d) CH₄O       Nonmetal-Nonmetal           3.44 - 2.55 = 0.89                Covalent

   Carbon, Hydrogen and oxygen are nonmetals

2) Silver coins can conduct electricity.

Taking into account the definition of ionic and covalent bond and conductive materials :

An ionic bond is a type of chemical bond that occurs when one atom gives up an electron to the other, in order for both to achieve electronic stability.

This union normally occurs between metal and nonmetal elements with different electronegativity, which means that the elements have different capacity to attract electrons.

In other words, an ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions. Usually, the metal gives up its electrons forming a cation to the nonmetal element, which forms an anion.

The covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule.

Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.

In this case, you know that:

Then, the compound:

a. Magnesium oxide (MgO) Ionic

b. Strontium chloride (SrCl₂) Ionic

c. Ozone (O₃) Covalent

d. Methanol (CH₄O) Covalent

Electrical conductivity is the property of a material that allows an electrical current to travel through its atomic structure, with low resistance from this material.

Conductive materials are those that offer little resistance to the passage of electricity. Electrons can circulate freely through material because they are loosely bound to atoms and can therefore conduct electricity.

In other words, conductive materials allow the free flow of electrons between particles, facilitating the conduction of electricity across the entire surface.

Conductors, then, are those that have a large number of free electrons that move through the material, transmitting charge more easily from one object to another.

Metals have several million atoms, each with two or three electrons in its outer orbit (valence electrons). These valence electrons, in metals, are characterized by a tendency to release electrons to achieve a certain stability in terms of their configuration. In this way they conduct electricity.

Silver Ag is a metal. Then a silver coin will conduct electricity.

Learn more about:

Answer:

Although it is possible to have more than one state, it is also possible to have only one state.

Explanation:

Answer:

55%

Explanation:

Let A represent isotope eu-151

Let B represent isotope eu-153

Let A% represent Abundance of isotope A (eu-151)

Let B represent abundance of isotope B (eu-153)

The abundance of isotope A (eu-151) can be obtained as follow:

Step 1:

Data obtained from the question include:

Atomic mass of Europium = 151.9 amu

Mass of isotope A (eu-151) = 151.0 amu

Mass of isotope B (eu-153) = 153.0

amu

Abundance of isotope A (eu-151) = A%

Abundance of B (eu-153) = B% = 100 - A%

Step 2:

Determination of the abundance of Abundance of isotope A (eu-151). This is illustrated below:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

151.9 = [(151 x A%)/100] + [(153x B%)/100]

151.9 = [(151 x A%)/100] + [(153x (100-A%))/100]

151.9 = [151A%/100] + [15300 -

153A%/100]

151.9 = (151A% + 15300 - 153A%) /100

Cross multiply to express in linear form

151.9 x100 = 151A% + 15300 - 153A%

15190 = 151A% + 15300 - 153A%

Collect like terms

15190 - 15300= 151A% - 153A%

- 110 = - 2A%

Divide both side by - 2

A% = - 110 / - 2

A% = 55%

Therefore the abundance of eu-151 is 55%

Answer:

The natural abundance of eu-153 is 45.0 %

The natural abundance of eu-151 is 55.0 %

Explanation:

Step 1: Data given

The atomic mass of europium is 151.9 amu.

eu-151 hass with a mass of 151.0 amu

eu153 hass a mass of 153.0 amu

Step 2: Calculate the percent natural abundance

natural abundance eu-151 = X

natural abundance eu-153 = 1-X

151.9 = 151.0* X + 153.0 * (1-X)

151.9 = 151.0X + 153.0 -153.0 X

-1.1 = -2.0 X

X = 0.55 = 55 %

The natural abundance of eu-153 is 45.0 %

The natural abundance of eu-151 is 55.0 %

Answer:

Option C is correct.

The minimum amount of material that is needed for a fission reaction to keep going is called the critical mass.

Explanation:

Nuclear fission is the term used to describe the breakdown of the nucleus of a parent isotope into daughter nuclei.

Normally, the initial energy supplied for nuclear fission is the energy to initiate the first breakdown of the first set of radioactive isotopes that breakdown. Once that happens, the energy released from the first breakdown is enough to drive further breakdown of numerous isotopas in a manner that leads to more energy generation.

But, for this to be able to be sustained and not fizzle out, a particular amount of radioactive material to undergo nuclear fission must be present. This particular amount is termed 'critical mass'

Hope this Helps!!!

At the end the energy of the system is the same as 200 J.

Explanation:

It is a closed complex system and it has 200 J of energy.

Here the system is converting from one form to another form and then to another form.

As we know that there is a conservation of energy when one form is converted to another form without losing or gaining energy, whereas it may convert from one form to another form with the energy being conserved.

So after there is a conversion of one form to another form but there is no loss or gain of energy and same energy must be retained.

Answer:

C Al = 0.8975 J/g.K

Explanation:

∴ m Al = 28.4 g

∴ T Al = 39.4°C = 312.4 K

∴ m H2O = 50.0 g

∴ T1 H2O = 21°C = 294 K

∴ T2 H2O = 23°C = 296 K

∴ C H2O = 4,18 J/g.K

⇒ C Al = ?

in a calorimeter:

∴ Al give heat: Q Al < 0

∴ H2O revceives heat: Q H2O > 0

⇒ - Q Al = Q H2O

⇒ - (28.4 g)*(C Al)*(296 K - 312.4 K) = (50.0 g)*(4.18 J/g.K)*(296 K - 294 K)

⇒ - (- 465.76 g.K)*(C Al) = 418 J

⇒ C Al = (418 J) / (465.76 g.K)

⇒ C Al = 0.8975 J/g.K = 897.5 J/Kg.K

The specific heat capacity of aluminum in this example is 0.394 J/g °C. This was determined by calculating the heat gained by water and considering that it equals the heat lost by the aluminum, and subsequently solving for the specific heat capacity of aluminum.

To find the specific heat capacity (C) of aluminum, we must consider the amount of heat transferred from the aluminum to the water (expressed as q). The heat gained by water is calculated using the equation q = m * C * ΔT, where m is the mass, C is the specific heat, and ΔT is the change in temperature. According to the question, water's mass (m) is 50.0g, its specific heat (C) is 4.184 J/g °C, and the difference in its temperature (ΔT) is 2.00 °C. So, the heat gained by water is q = 50.0g * 4.184 J/g * °C * 2.00 °C = 418.4 J.

The heat lost by aluminum is equal to the heat gained by water. Therefore, using the equation q = m * C * ΔT and plug in the values of q (418.4 J), m (28.4 g), and ΔT (39.4 °C), to solve for C, the specific heat of aluminum, we can rearrange the formula to C = q / (m * ΔT) = 418.4 J / (28.4g * 39.4 °C) = 0.394 J/g °C.

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Answer:

4.2 L O₂ is needed to completely react with 2.8 L hydrogen sulfied.

Explanation:

Without pressure and temperature we cannot calculate the this vale

We assume that the reaction  take place under standard Temperature and Pressure(STP).

At STP, One mole ([tex]6.023\times 10^{23}[/tex] particles) of any gas occupied volume 22.4 L.

The balanced equation of this reaction is

[tex]2H_2S+3O_2\rightarrow 2SO_2+2H_2O[/tex]

Now we use molar ratio.

[tex]2.8L\ H_2S . \ \frac{1 mol\ H_2S}{22.4L \ H_2S}\ . \ \frac{3 mol\ O_2}{2 mol \ H_2S} \ . \ \frac{22.4L\ O_2}{1 mol\ O_2}[/tex]

=4.2 L O₂

4.2 L O₂ is needed to completely react with 2.8 L hydrogen sulfied.

Final answer:

To determine the volume of oxygen needed to react with 2.8 liters of hydrogen sulfide, the stoichiometric ratio from the chemical equation is used, revealing that 4.2 liters of oxygen is required for complete reaction.

Explanation:

Stoichiometry of Hydrogen Sulfide and Oxygen Reaction

The question involves the stoichiometric relationship between hydrogen sulfide and oxygen gases during a chemical reaction. The balanced chemical equation for the reaction between hydrogen sulfide (H₂S) and oxygen (O₂) to produce sulfur dioxide (SO₂) and water (H₂O) is:

2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(g)

To find the volume of oxygen needed to react with 2.8 liters of hydrogen sulfide, we use the stoichiometric coefficients from the balanced equation, which tell us that 2 volumes of H₂S react with 3 volumes of O₂.

This gives us a ratio of:

2H₂S : 3O₂

Using this ratio, we can find the volume of oxygen required by setting up the proportion:

(2.8 L H₂S) / (2L H₂S) = (x L O₂) / (3L O₂)

Solving for x, we get:

x = (2.8 L H₂S) × (3L O₂) / (2L H₂S)

x = 4.2 L O₂

Therefore, 4.2 liters of oxygen gas is needed to completely react with 2.8 liters of hydrogen sulfide gas.

Answer:

Concentration of hydrogen ions in solution

Brainliest would be appreciated

Final answer:

pH measures the concentration of hydrogen ions in a solution and indicates its acidity or basicity. The pH scale ranges from 0 to 14, with lower values being acidic and higher values alkaline. The pH value represents the negative logarithm of the hydrogen ion concentration.

Explanation:

pH measures the concentration of hydrogen ions (H+) in a solution. In pure water, a small percentage of water molecules dissociate into equal numbers of hydrogen ions (H+) and hydroxide ions (OH−), which is a source of hydrogen ions in solutions. The pH scale, which ranges from 0 to 14, determines a solution's acidity or basicity; a pH less than 7 indicates acidity, greater than 7 indicates alkalinity, and a pH of 7 is neutral. The pH value is a negative logarithm of the hydrogen ion concentration, meaning that each whole number on the pH scale represents a tenfold increase or decrease in hydrogen ion concentration.

The pH of a solution provides a quick way to determine its acidity or basicity. High concentrations of hydrogen ions yield a low pH, whereas low concentrations result in a high pH. The relative acidity or alkalinity of a solution can significantly impact various chemical and biological processes, making pH an essential parameter in science and various industries.

Answer:

Rocks can be simply defined as the aggregates of various minerals, and these broken rock fragments and mineral grains are formed from the weathering of rocks. Weathering refers to the breakdown of rocks due to the occurrence of various geological processes that are initiated by the agents such as wind, water and ice.

The weathering process leads to the formation of sediments of variable size and shape, and these are such as pebbles, sand, silt and clay particles.

Answer:

1.   67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

[tex]In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})[/tex]

Given that:

time [tex]t_1[/tex] = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time [tex]t_2[/tex] = 10.6 hours

Temperature [tex]T_1[/tex] = 0° C = (0+273 )K = 273 K

Temperature [tex]T_2[/tex] = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since [tex](\frac{k_2}{k_1}=\frac{t_2}{t_1})[/tex]

Then we can rewrite the above expression as:

[tex]In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})[/tex]

[tex]In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})[/tex]

[tex]2.934 = \frac{E_a}{8.314}(\frac{30}{82719})[/tex]

[tex]2.934 = \frac{30E_a}{687725.766}[/tex]

[tex]30E_a = 2.934 *687725.766[/tex]

[tex]E_a = \frac{2.934 *687725.766}{30}[/tex]

[tex]E_a =67255.58 \ J/mol[/tex]

[tex]E_a =67.2 \ kJ/mol[/tex]

Answer: [tex]KCrO_4[/tex]

Explanation:

First, calculate how much percent of oxygen there is. We know that the whole compound cannot exceed 100%, so we take that, and substract it from 26.56% and 35.41%.

 100.00

-   26.56

-------------------

   73.44

-   35.41

---------------------

   38.03

That is how much of oxygen we have.

Potassium: 26.56% or 26.56g

Chromium: 35.41% or 35.41g

Oxygen: 38.03% or 38.03g

To find the empirical formula, you simply find the amount of mol that each one has. You can do this by using the atomic mass of each element.

[tex]Potassium: 26.56g(\frac{1mol}{39.10g})= 0.67mol[/tex]

[tex]Chromium: 35.41g(\frac{1mol}{52g})=0.68mol[/tex]

[tex]Oxygen: 38.03g(\frac{1mol}{16g})=2.38mol[/tex]

We now determine the lowest number and divide each mol by it. In this case, the lowest number is 0.67

[tex]Potassium: \frac{0.67mol}{0.67}=1[/tex]

[tex]Chromium:\frac{0.68mol}{0.67}=1[/tex]

[tex]Oxygen:\frac{2.38mol}{0.67}=3.55 = 4[/tex]

Finally, we take each element add add their respective number.

So, this empirical formula would be:

[tex]KCrO_4[/tex]

The empirical formula of a compound with 26.56% potassium, 35.41% chromium, and the rest being oxygen would be [tex]K_2Cr_2O_7[/tex]

The number of mole of each element in the compound can be found by dividing each element's percentage with their respective molar weights:

Potassium K = 26.56%

               = 26.56/39.1

                    = 0.68

Chromium, Cr = 35.41%

                       = 35.41/52

                        = 0.68

Oxygen, O = 100 - 26.56+35.41

                     = 38.03/16

                        = 2.38

Divide each number of moles by the smallest.

K = 0.68/0.68

         = 1

Cr = 0.68/0.68

       = 1

O = 2.38/0.68

      = 3.5

Thus, the empirical formula would be [tex]KCrO_{3.5}[/tex]

Multiply all by 2 to remove the fraction:

[tex]K_2Cr_2O_7[/tex]

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Answer:

[tex]T_2=571.9K[/tex]

Explanation:

Hello,

In this case, we consider the following formula defining the energy and the temperature change for the sample of iron:

[tex]Q=n_{Fe}Cp_{Fe}(T_2-T_1)[/tex]

Now, solving the final temperature, considering a positive inlet heat, we have:

[tex]T_2=T_1+\frac{Q}{n_{Fe}Cp{Fe}} =293K+\frac{3500J}{0.5mol*25.1J/(mol*K)} \\T_2=571.9K[/tex]

Best regards.

Answer:

Explanation:

Step 1: Data given

Number of heat transfer = 3500 J

Number of moles of iron = 0.5 moles

Initial temperature = 293 K

The molar heat of iron is 25.1 J/(mol*K)

Step 2: Calculate ΔT

Q = n* C * ΔT

⇒with Q = the heat transfer = 3500 J of energy

⇒with n = the number of moles iron  = 0.5 moles

⇒with C = the molar heat of iron = 25.1 J/mol*K

⇒ΔT = the change of temperature = T2 - T1 = T2 - 293 K

3500 J = 0.5 moles *25.1 J/mol * K * ΔT

ΔT = 278.9

Step 3: Calculate ΔT

ΔT = 278.9 = T2 - T1 = T2 - 293 K

T2 = 278.9 + 293 K

T2 = 551.9 K

Answer:

The ideal gas law ... A gas turbine, which uses continuous combustion, simply exhausts its ... This makes them ideal for use in vehicles, as they also start up more ... A four stroke engine delivers one power stroke for every two cycles of ... ignition, exhaust) however, these steps occur 3 times per one spin of. Internal combustion engine.

Explanation:

Ideal gas law is applicable  to each of the steps of the 4 stroke engine.

The ideal gas law is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.

It was first stated by Benoît Paul Émile Clapeyron in 1834 as a combination of the empirical Boyle's law, Charles's law, Avogadro's law, and Gay-Lussac's law. The ideal gas law is often written in an empirical form.

The state of an amount of gas is determined by its pressure, volume, and temperature. The modern form of the equation relates these simply in two main forms. The temperature used in the equation of state is an absolute temperature: the appropriate SI unit is the kelvin.

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7 moles of BaBr₂ produces 14 moles of KBr.

Explanation:

Given reaction is

BaBr₂ + K₂SO₄ → KBr + BaSO₄

It can be balanced by putting 2 in front of KBr as,

BaBr₂ + K₂SO₄ → 2 KBr + BaSO₄

From the above balanced equation, it was found that, 1 mole of Barium bromide required to produce 2 moles of KBr.

Now the molar ratio of BaBr₂ to KBr is written as 1 : 2.

In the same way, the molar ratio for 7 moles of BaBr₂, can be written as 7:14.

So 7 moles of BaBr₂ produces 14 moles of KBr.

The moles of KBr formed has been 14 mol.

The balanced chemical equation for the reaction has been:

[tex]\rm BaBr_2\;+\;K_2SO_4\;\rightarrow\;2\;KBr\;+\;BaSO_4[/tex]

From the balanced chemical equation, the coefficient has been describing the moles of reactants and products formed.

The balanced chemical equation has been given as:

[tex]\rm 1\;mol\;BaBr_2=2\;mol\;KBr[/tex]

The moles of KBr formed by 7 moles of Barium bromide has been as:

[tex]\rm 1\;mol\;BaBr_2=2\;mol\;KBr\\7 \;mol\;BaBr_2=2\;\times\;7 \;mol\;KBr\\7 \;mol\;BaBr_2=14 \;mol\;KBr\\[/tex]

The moles of KBr formed has been 14 mol.

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Answer:

Polar Jet stream

Explanation:

Polar Jet stream is also called the polar front jet or mid latitude jet steam, it is a very powerful belt of the upper level winds which sits above the polar front. It is the strongest wind in the tropopause, it's movement is towards the westerly direction of the mid latitude.

Answer:

40g of CO2 at 10C

Explanation:

Since average kinetic energy depends on absolute temperature (directly proportional to absolute temperature) and independent of amount and nature of gas. given that is have same temperature.

40g of CO2 at 10C

Any gas at 10 ℃ would have the same average kinetic energy as 10.0 grams of CO2 at the same temperature due to the principles of the kinetic molecular theory.

The conditions that contain molecules with the same average kinetic energy as the molecules in 10.0 grams of CO2 at 10 ℃ would be any other mass of gas at the same temperature, since the average kinetic energy of the molecules of a gas depends only on temperature according to the kinetic molecular theory. This is true regardless of the type of gas or its mass, as long as the gases are at the same temperature.

For example, if you have helium gas at 10 ℃, the average kinetic energy of its molecules would be the same as that of the CO2 molecules at 10 ℃. This is because the kinetic molecular theory posits that all gases at the same temperature have the same average kinetic energy.

Answer:

7.1 × 10⁹ years (D)

Explanation:

half life of the Uranium-235 = 710 million years

to find the time it will take for the U-235 need to be stored securely to be safe

taken the full percent = 100

100 / 2ⁿ = 0.1 where n is number of half-life it has undergone

100 / 0.1 = 2ⁿ

1000 = 2ⁿ

take log of both side

log 1000 / log 2 = n

n = 9.967 number of half-lives

the number of years it will take = 710 million × 9.967 number of half-lives = 7075.7 × 10⁶ years approx 7.1 × 10⁹ years

The correct option is D. [tex]7.1 \times 10^9 years[/tex]

[tex]100 \div 2^n = 0.1[/tex]

here n is number of half-lives it has undergone

[tex]100 \div 0.1 = 2^n\\\\1000 = 2^n[/tex]

Now

take log of both side

[tex]log 1000 \div log 2[/tex] = n

n = 9.967 number of half-lives

Now

the number of years it will take should be

[tex]= 710\ million \times 9.967 \\\\= 7075.7 \times 10^6 \\\\= 7.1 \times 10^9 years[/tex]

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Answer:

Approx. 8 grams of KNO3 will not dissolve

Explanation:

According to the curve at 20 degrees C only 32 grams of KNO3 can dissolve in 100 grams of water meaning if you hvae 40 grams of KNO3 in water at 20 degrees C ; 40-32= 8