Which compound in the picture has a central atom with four outside atoms and has at least one nonpolar bond.

Answer:

0.74 atm.

Explanation:

From the question given above, the following data were obtained:

Pressure of Ne (Pₙₑ) = 1.52 atm

Pressure of He (Pₕₑ) = 766 mmHg

Total pressure (Pₜ) = 3.27 atm

Pressure of Ar (Pₐᵣ) =?

Next, we shall convert the pressure of He from mmHg to atm. This can be obtained as follow:

760 mmHg = 1 atm

Therefore,

766 mmHg = 766 mmHg × 1 atm / 760 mmHg

766 mmHg = 1.01 atm

Finally, we shall determine the partial pressure of Ar. This can be obtained as follow:

Pressure of Ne (Pₙₑ) = 1.52 atm

Pressure of He (Pₕₑ) = 1.01 atm

Total pressure (Pₜ) = 3.27 atm

Pressure of Ar (Pₐᵣ) =?

Pₜ = Pₙₑ + Pₕₑ + Pₐᵣ

3.27 = 1.52 + 1.01 + Pₐᵣ

3.27 = 2.53 + Pₐᵣ

Collect like terms

3.27 – 2.53 = Pₐᵣ

Pₐᵣ = 0.74 atm

Thus the partial pressure of Ar is 0.74 atm.

Answer:

where are the statements??

Three moles of methane have a volume of 67.2 L.

The area occupied by gaseous particles under normal temperature and pressure circumstances is referred to as the volume of gas. It is identified as a "V." The letter "L" stands for "liters," the SI unit of volume. At normal temperature, a mole of gas has a volume of 24 \(m^{3}\), or 24 000 \(cm^{3}\).

While gases lack a distinct structure, unlike solids and liquids, they also lack a distinct volume. Typically, a substance's volume is not considerably altered when it transitions from solid to liquid.

3 mole Equals 1 mole of methane gas.

Given that, at specific temperatures and pressures, 1 mole of any gas takes up 22.4 L of space.

The volume that 3 mole of meta he gas takes up will then be

3 × 22.4 = 67.2L

67.2 L is the volume of 3 moles of methane.

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De acuerdo con la imagen, podemos inferir que el vapor de etanol está ejerciendo presión sobre el agua que está en el conducto causando que esta se desplace.

En la imagen se ven dos recipientes con etanol líquido. Una de ellas está experimentando un aumento de calor que causa que las moléculas de etanol comiencen a moverse más rápido. Adicionalmente, las convierten en etanol gaseoso. Estas moléculas de etanol gaseoso hacen que el agua que está contenida en el conducto del recipiente se desplace como resultado de la presión del etanol.

English Version

Based on the image, we can infer that the ethanol vapor is exerting pressure on the water that is in the conduit, causing it to move.

What is seen in the image?

In the image you can see two containers with liquid ethanol. One of them is experiencing a rise in heat that causes the ethanol molecules to start moving faster. Additionally, the configurations in gaseous ethanol. These gaseous ethanol molecules cause the water that is contained in the conduit of the container to move as a result of the pressure of the ethanol.

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The atomic hybrid orbitals which make up the bond between C and O in formaldehyde, \(CH_{2} O\) will be one s-orbital and two p orbital make sp2 hybridization.

The 3 new hybrid orbitals can only be created through  \(sp^{2}\) hybridization, which occurs whenever an atom's two pi orbitals including one s orbital overlap with mix together. Consequently, \(CH_{2} O\) (Formaldehyde) has \(sp^{2}\)hybridization.

It is found that when atoms share one electron then it will form sigma bond whereas when atoms share more than one electrons then it will form pi bond.

Therefore, the hybridization of  \(CH_{2} O\)  will be  \(sp^{2}\) hybridization.

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Answer:

0.1 moles of N2 will react with 0.3034 moles of Mg.

Explanation:

\(\sf{balanced \ eqaution = 3Mg + N2 \rightarrow Mg3N2}\)

using molar ratio:

3Mg : N2

3 : 1

moles of N2:

\(\hookrightarrow \sf\frac{0.3034 }{3} *1\)

\(\hookrightarrow \sf 0.10113 \ moles \ of \ N2\)

Answer:

46.07

Explanation:

im not 100 sure this is right almost tho

please mark brainliest

Answer:

B) They will react because X and Y can share two pairs of electrons to become stable

Explanation:

The electron configurations of two elements x and y are given :

X: 1s2 2s2 2p6

Y: 1s2 2s2 2p6 3s2 3p6

The statement that is true for both the elements is that, they both will react as they both can share two pairs of electrons to become stable.

To become stable the outermost shell or p orbital should have 8 electrons, so element X  can gain 2 atoms to become stable.

Element Y can also react as it can also share two atoms to fulfill its 3p orbital and will stable.

Hence, the correct option is "B".

The given representation of one unit of C₆H₁₂O₆ in water is incomplete as it does not include the water molecules that are essential for the dissolution process.

In the given representation, only the C₆H₁₂O₆ molecules are shown, while the water molecules are intentionally not depicted. However, when C₆H₁₂O₆  dissolves in water, it forms a solution where C₆H₁₂O₆ molecules are surrounded by water molecules, resulting in a hydrated state.

Therefore, the representation is incomplete and inaccurate since it neglects the presence of water molecules, which play a crucial role in the dissolution and formation of a hydrated C₆H₁₂O₆ complex in water.

The question should be:

A representation of one unit of C₆H₁₂O₆ in water is shown below. (The water 12 molecules are intentionally not shown.)

(a) What is wrong with this representation?

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Answer:

yo no se de eso preguntale a otro si

Answer:

6

Explanation:

if there is 10n and the friction is 4n on the table then it is 6n.

i hope this helps you! :)

If hydrochloric acid is added to a solution of HF, it will react with the HF to form H3O+ and F- ions. This reaction is a type of acid-base reaction, where the hydrochloric acid acts as the stronger acid and donates a proton to the HF, which acts as the weaker base.

The addition of hydrochloric acid will increase the concentration of H3O+ ions in the solution. This increase in the concentration of H3O+ ions will shift the equilibrium of the HF ionization reaction to the left, leading to a decrease in the percent ionization of HF. This is because the increased concentration of H3O+ ions will make it more difficult for the HF molecules to ionize and release H+ ions into the solution.

Therefore, the correct answer is that the percent ionization of HF will decrease. The Ka for HF will not change, as it is a constant value for a given temperature and is independent of the concentration of H3O+ ions in the solution.

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The percent yield of \(ClF_3\) was obtained is 39.97% when the reaction of 7.1 grams of chlorine with excess fluorine produced 7.4 grams of \(ClF_3\).

Given the mass of chlorine = 7.1g

The mass of chlorine trifluoride produced = 7.4g

Molar mass of chlorine used = 70.906g/mole

Molar mass of \(ClF_3\) used = 92.448g/mole

The reaction of chlorine with fluorine to produce chlorine tri fluoride is as:

\(Cl_2 + 3F_2 -- > 2ClF_3\)

1 mole of \(Cl_2\) is used to produce 2 moles of \(ClF_3\)

mass of \(Cl_2\) used = number of moles x molar mass = 1 * 70.906= 70.906g

Mass of \(ClF_3\) used = 2 * 92.448 = 184.896g

Here for 70.906g of  \(Cl_2\), 184.896g of \(ClF_3\) is produced.

Then for 7.1g of \(Cl_2\) = 184.896 * 7.1/70.906 = 18.51g of  is produced.

Theoretical Yield of \(ClF_3\) = 18.51g

Actual Yield = 7.4g of \(ClF_3\)

Percent Yield =\((7.4/18.51 )*100 = 39.97\)% yield of \(ClF_3\)

Therefore, 39.97% yield of \(ClF_3\) was obtained.

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Vascular and nonvascular organisms in the plant kingdom use photosynthesis to get nutrients.

All plants make use of photosynthesis to get food.

Vascular plants have vascular tissues such as xylem, phloem, etc. while non-vascular plants are devoid of vascular tissues.

Vascular tissues are used to conduct water (xylem) and manufactured foods (phloem) around the body of plants.

Lower plants have other means (such as diffusion and pseudo-vascular tissues) to conduct materials around their own bodies.

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A buffer is made from an acid - base conjugate pair as shown by option A

Buffers are often composed of weak acids and their conjugate bases (or weak bases and their conjugate acids). The weak acid can contribute a proton to balance any new base, whereas the conjugate base can absorb a proton to do so.

This balance between the acid and its conjugate base allows the buffer to survive pH changes. Buffers are essential in biological systems because many biochemical processes are particularly sensitive to pH changes.

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When 50 NBr3 molecules and 57 NaOH formula units react according to the given balanced equation, the result is the formation of 150 NaBr formula units.

According to the balanced equation provided:

2 NBr3 + 3 NaOH -> N2 + 3 NaBr + 3 HOBr

From the equation, we can see that 2 moles of NBr3 react with 3 moles of NaOH to form 3 moles of NaBr.

To determine the number of NaBr formula units formed, we need to convert the given quantities into moles.

Given:

Number of NBr3 molecules = 50

Number of NaOH formula units = 57

To convert the number of NBr3 molecules to moles, we need to divide the given quantity by Avogadro's number. Similarly, for NaOH formula units, we can directly consider them as moles.

Using Avogadro's number (6.022 x 10^23 molecules/mol), we can calculate the number of moles for NBr3 and NaOH:

Number of moles of NBr3 = 50 / (6.022 x 10^23)

Number of moles of NaOH = 57

Now, we can use the mole ratios from the balanced equation to determine the number of moles of NaBr formed. From the equation, we know that 2 moles of NBr3 react to form 3 moles of NaBr.

Number of moles of NaBr = (Number of moles of NBr3) x (3 moles of NaBr / 2 moles of NBr3)

Finally, we can convert the number of moles of NaBr to the number of NaBr formula units using Avogadro's number:

Number of NaBr formula units = (Number of moles of NaBr) x Avogadro's number

Calculating these values, we find that 50 NBr3 molecules and 57 NaOH formula units react to form 150 NaBr formula units.

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A.2-octene  is the name of the molecule shown below.

The oct part of the name indicates the parent chain. In this case, oct- is an abbreviation for octane, which means that there are 8 carbons in the parent chain. The first part of the name, cis-2. 2 is used to indicate the first atom that forms the double bond above.

Alkenes are classified into higher olefins and alpha olefins. This means that the double bond is in the alpha (primary) position, which makes the compound more reactive and provides useful chemistry.

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The reaction occurs in The anode of an electrochemical cell. Option B is the correct option

This refers to a device that can produce electrical energy from the chemical reactions occurring in it, or use the electrical energy to facilitate chemical reactions. ​

An electrochemical cell splits the oxidant and reductant in a way that allows electrons to flow through an external circuit from the reductant (which gets oxidized) to the oxidant (which causes reduction) while preventing them from physically touching each other.

The two compartments of an electrochemical cell where the half-reactions occur are called the anode(Oxidation occurs here) and the cathode (Reduction occurs here), and they must have an electrode that you can connect to the external circuit.

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In talc preparation, it is used during the filtration process and also acts as a distribution agent. When it is added to the dill water, it combines with all other compounds present in the substance and leaves the water alone. In this way, it acts as a filtering agent.

500 ml of a solution with an ultimate concentration of 2.7M was made using 75.0 ml of 18.0 m h2so4.

Molarity is defined as the number of moles of a solute dissolved in one liter of the solution. It is denoted by ‘M’ and its unit is moles per liter (mol/L). Formula to calculate Molarity (M): Molarity = Number of moles of solute / Volume of solution (in liters)

Given data: Initial volume of H2SO4 = 75.0 mLInitial molarity of H2SO4 = 18.0 MVolume of a solution prepared = 500 mL = 0.5 LC. converting initial volume of H2SO4 to liters:Initial volume of H2SO4 = 75.0 mL = 75.0/1000 = 0.075 L. Number of moles of H2SO4 used for preparing the solution: Number of moles of H2SO4 = Molarity × Volume.

A numberr of moles of H2SO4 = 18.0 M × 0.075 L. Number of moles of H2SO4 = 1.35 MGiven that the volume of the solution prepared = 500 mL = 0.5 L.

The final molarity of H2SO4 in the prepared solution is given by:Final molarity of H2SO4 = Number of moles of H2SO4 / Volume of solution. Final molarity of H2SO4 = 1.35 M / 0.5 L. Final molarity of H2SO4 = 2.7 M .Therefore, the final concentration of the solution is 2.7 M.

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Answer:

4.00 moles of water

Explanation:

The equation of the reaction is:

2 C2H2Cl (9) + 5O2(g) -------> 4CO2 (g) + 2 H20( g) + 2 HCl(g)

To find the limiting reactant,  we find the species that yields the lowest number of moles of water.

From the reaction equation;

2 moles of C2H2Cl  yields 2 moles of water

10 moles of C2H2Cl  yields 10 * 2/2 = 10 moles of water

Similarly;

5 moles of oxygen yields 2 moles of water

10 moles of oxygen will yield 10 *2/5 =4 moles of water

Hence oxygen is the limiting reactant.

10 moles of oxygen yields 4 moles of water.

Answer:

After three hours, concentration of C₂F₄ is 0.00208M

Explanation:

As the units of the rate constant of the reaction are M⁻¹s⁻¹ we can know this reaction is of second order. Its integrated law is:

\(\frac{1}{[A]} =\frac{1}{[A]_0} +Kt\)

Where [A] and [A]₀ represents initial and final concentrations of the reactant (C₂F₄), K is rate constant (0.0410M⁻¹s⁻¹) and t is time.

3.00 hours are in seconds:

3 hours ₓ (3600 seconds / 1 hour) = 10800 seconds

Computing in the equation:

\(\frac{1}{[A]} =\frac{1}{[0.105mol / 4L]} +0.0410M^{-1}s^{-1}*10800s\\\frac{1}{[A]} = 480.9M^{-1}\\\\)

[A] = 0.00208M

The energy of combustion per gram of vanillin is 158.61 kJ/g and per mole of vanillin is 1.044 kJ/mol.

Given data:

The energy released by combustion of benzoic acid is 26.42 kJ/g. The combustion of 0.1584 g benzoic acid increases the temperature of a bomb calorimeter by 2.54 °C. The heat capacity of the calorimeter needs to be calculated. A 0.2130 g sample of vanillin (C8H8O3) is then burned in the same calorimeter, and the temperature increases by 3.25 °C.

To calculate the heat capacity of the calorimeter, we will use the following formula:

Q = CΔT where, Q = Heat absorbed by the calorimeter ,C = Specific heat capacity of the calorimeter ,

ΔT = Change in temperature

We are given: Q = Heat absorbed by the calorimeter = Energy released during combustion of benzoic acid = 26.42 kJ/gΔT = Change in temperature = 2.54 °C = 2.54 K

And we need to find C.

Substituting the values in the formula, we get: 26.42 kJ/g = C × 2.54 KC = 26.42/2.54 = 10.39 kJ/K

Now, we need to calculate the energy of combustion per gram of vanillin.To calculate the energy of combustion per gram of vanillin, we will use the following formula:

Energy of combustion per gram of vanillin = Q/m

where, Q = Heat absorbed by the calorimeter  , m = Mass of the sample of vanillin burned

3.25 °C = 3.25 K

Mass of the sample of vanillin burned = 0.2130 g

Energy released during combustion of vanillin:ΔT = 3.25 K = Change in temperature

C = 10.39 kJ/K = Specific heat capacity of the calorimeter

Q = CΔT = 10.39 kJ/K × 3.25 K = 33.76 kJ

The energy released during combustion of vanillin:

Energy of combustion per gram of vanillin = Q/m= 33.76 kJ/0.2130 g= 158.61 kJ/g

Now, we need to calculate the energy of combustion per mole of vanillin.

To calculate the energy of combustion per mole of vanillin, we will use the following formula:

Energy of combustion per mole of vanillin = Energy of combustion per gram of vanillin/Molecular weight of vanillin

We are given:

Energy of combustion per gram of vanillin = 158.61 kJ/g

Molecular weight of vanillin = 8 × 12 + 8 × 1 + 3 × 16 = 152 g/mol

Energy of combustion per mole of vanillin:

Energy of combustion per mole of vanillin = Energy of combustion per gram of vanillin/Molecular weight of vanillin= 158.61 kJ/g / 152 g/mol= 1.044 kJ/mol

Hence, the energy of combustion per gram of vanillin is 158.61 kJ/g and per mole of vanillin is 1.044 kJ/mol.

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Answer:The process of photosynthesis in plants converts carbon dioxide and water into glucose and oxygen. The balanced equation for this reaction is:

6 CO2 + 6 H2O --> C6H12O6 + 6 O2

This equation tells us that for every 6 molecules of carbon dioxide (CO2) that are consumed, 1 molecule of glucose (C6H12O6) is produced, along with 6 molecules of oxygen (O2).

Given that we have 760.0 grams of carbon dioxide and a sufficient amount of water, we can use the balanced equation and the molar mass of glucose to calculate the amount of glucose that will be produced.

First, we need to convert the given amount of carbon dioxide to moles. We can do this by using the molar mass of carbon dioxide, which is 44.01 g/mol.

760.0 g CO2 / 44.01 g/mol = 17.3 moles CO2

Next, we can use the balanced equation to determine the number of moles of glucose that will be produced. Since the ratio of CO2 to glucose is 6:1, for every 6 moles of CO2, 1 mole of glucose will be produced.

17.3 moles CO2 / 6 = 2.88 moles glucose

Finally, we can convert the number of moles of glucose to grams using the molar mass of glucose, which is 180.16 g/mol

2.88 moles glucose * 180.16 g/mol = 517.3 grams glucose

Therefore, plants will produce 517.3 grams of glucose using 760.0 grams of carbon dioxide and a sufficient amount of water.

Explanation: