If 0.4743 moles of H2O are produced, how many grams of VOCl3 will also be produced?
(V2O5 + 6 HCl → 2 VOCl3 + 3 H2O)

Answer: The answer is 5.15x10^22

Explanation:

The formula unit present in a teaspoon of salt [tex]NaCl[/tex] having a mass of about 5.0 g is [tex]5.15 \times10^{22}[/tex] formula units.

Molar mass, also known as molecular weight, is the mass of one mole of a substance. It is calculated by summing up the atomic masses of all the atoms in a molecule. The unit of molar mass is grams per mole (g/mol).

Now, to determine the number of formula units in a teaspoon of salt (NaCl), we need to use Avogadro's number and the molar mass of NaCl.

Avogadro's number [tex](N_a)[/tex] is approximately. [tex]6.022 \times10^{23}[/tex] formula units per mole.

The molar mass of [tex]NaCl[/tex] is the sum of the atomic masses of sodium (Na) and chlorine ([tex]Cl[/tex]), which are approximately 22.99 g/mol and 35.45 g/mol, respectively.

To calculate the number of formula units in 5.0 g of [tex]NaCl[/tex], we can follow these steps:

Now, calculate the number of moles of [tex]NaCl[/tex] using its molar mass:

Moles = Mass / Molar mass

Moles = [tex]5.0 g[/tex] / [tex](22.99 g/mol + 35.45 g/mol)[/tex]

Calculate the number of formula units using Avogadro's number:

Formula units = [tex]Moles \times Avogadro's number[/tex]

Let's perform the calculation:

Molar mass of [tex]NaCl[/tex]= [tex]22.99 g/mol + 35.45 g/mol = 58.44 g/mol[/tex]

Moles of [tex]NaCl[/tex] = [tex]5.0 g[/tex] / [tex]58.44 g/mol[/tex] ≈ [tex]0.0856 mol[/tex]

Formula units = [tex]0.0856 mol \times (6.022 \times 10^{23})[/tex] formula units/mol ≈ [tex]5.15 \times10^{22}[/tex]formula units.

Therefore, there are approximately [tex]5.15 \times10^{22}[/tex] formula units in a teaspoon of salt ([tex]NaCl[/tex]) having mass [tex]5.0 g[/tex].

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Answer:

6.25e-6 is the value of the equilibrium constant

Explanation:

we have this equation

[tex]PbBr(s) ----- Pb^{2+}(aq) + 2Br(aq)[/tex]

When at a state of equilibrium,

we have the concentration of Pb^2+ to be 0.01

we have the concentration of Br^- to be 0.025

the equilibrium constant concentration of both pure solids and liquid s are said to be equal to 1

[PbBR2] = 1

such tht

Keq = [Pb^2+] x [Br-]^2

we already know the values of these from the above.

0.01x0.025^2

= 0.01 x 0.000625

= 0.00000625

= 6.25 x 10^-6

= 6.25e^-6

The heat of vaporization of benzene is required.

The heat of vaporization of benzene is 33009 J/kg.

[tex]T_0[/tex] = Normal boiling point = 80.1+273.15 K

[tex]T_B[/tex] = Boiling point at given pressure = 26.1+273.15 K

[tex]R[/tex] = Gas constant = 8.314 J/mol K

[tex]P[/tex] = Pressure at given [tex]T_B[/tex] = 100 torr

[tex]\Delta H[/tex] = Heat of vaporization

From the Clausius–Clapeyron equation

[tex]\dfrac{1}{T_B}=\dfrac{1}{T_0}-\dfrac{R\ln(\dfrac{P}{P_0})}{\Delta H}\\\Rightarrow \Delta H=\dfrac{R\ln\dfrac{P}{P_0}}{\dfrac{1}{T_0}-\dfrac{1}{T_B}}\\\Rightarrow \Delta H=\dfrac{8.314\times \ln\left(\frac{100}{760}\right)}{\frac{1}{80.1+273.15}-\frac{1}{26.1+273.15}}\\\Rightarrow \Delta H=33008.99\ \text{J/kg}[/tex]

The heat of vaporization of benzene is 33009 J/kg.

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Answer:I think it would be the third one.

Explanation:

Answer:

The Group 2 metals become more reactive towards the water as you go down the Group.

Explanation:

These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.

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Answer:

Explanation:

Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .

Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .

Answer:

A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.

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Answer:

B

Explanation:

I did the question before and got it right.

Answer:

Percent error = 3.7%

Explanation:

Given data:

Density of Al cylinder = ?

Weight of cylinder = 18 g

Diameter = 1.3 cm

Height = 5.2 cm

Actual density of Al = 2.7 g/cm³

Percent error = ?

Solution:

First of all we will calculate the volume of cylinder through given formula.

V = πr²h

r = diameter /2

V = 22/7 × (0.65 cm)²× 5.2 cm

V = 22/7 × 0.4225cm²× 5.2 cm

V = 6.89 cm³

Now we will calculate the density.

d = m/v

d = 18 g/ 6.89 cm³

d = 2.6 g/cm³

Percent error:

Percent error = measured value - actual value /actual value × 100

Percent error = 2.6g/cm³ - 2.7g/cm³ /2.7g/cm³  × 100

Percent error = 3.7%

Negative sign shows that measured or experimental value is less than actual value.

Answer: The answer is D.

Explanation:

The form of energy associated with the motion, position, or shape of an object is called mechanical energy. An object's mechanical energy is a combination of its potential energy and its kinetic energy. The basketball has both potential energy and kinetic energy.

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Answer:

Explanation:

Caffeine is a weak base with pKb = 10.4

Kb = 10⁻¹⁰°⁴ = 3.98 x 10⁻¹¹

molecular weight of caffeine = 194.2

455 x 10⁻³ g / L = 455 x 10⁻³ / 194.2 moles / L

concentration of given solution a = 2.343 x 10⁻³ M

Let the caffeine be represented by B .

B    +   H₂O =  BH + OH⁻

a - x                   x        x  

x² / ( a - x ) = Kb

x² / ( a - x ) = 3.98 x 10⁻¹¹

x is far less than a so a -x is almost equal to a

x² = 3.98 x 10⁻¹¹ x 2.343 x 10⁻³ = 9.32  x 10⁻¹⁴

x = 3.05 x 10⁻⁷

[ OH⁻ ] = 3.05 x 10⁻⁷

pOH = - log ( 3.05 x 10⁻⁷ )

= 7 - log 3.05

= 7 - 0.484 = 6.5

pH = 14 - 6.5 = 7.5  

The pH of 455 mg/L of caffeine is 7.5

Using the formula;

Mass concentration = molar concentration × molar mass

Molar mass of caffeine = 194 g/mol

Mass concentration of caffeine = 455 mg/L

Molar concentration = Mass concentration/molar mass

Molar concentration = 455 × 10^-3g/L/194 g/mol

= 0.00235 M

Let Caffeine by depicted by the general formula BH

We can now set up the ICE table as follows;

           :B + H2O   ⇄ BH       +       OH^-

I      0.00235              0                      0

C           - x                  +x                    +x

E    0.00235  - x           x                      x

Note that water is present in large excess

Again; pKb of caffeine =10.4

Kb = Antilog[-pKb]

Kb = Antilog [-10.4]

Kb = 3.98 × 10^-11

Kb = [BH] [OH^-]/[:B]

3.98 × 10^-11 = [x] [x]/[ 0.00235  - x  ]

3.98 × 10^-11 [ 0.00235  - x  ] = [x] [x]

9.4 × 10^-14 - 3.98 × 10^-11x = x^2

x^2 + 3.98 × 10^-11x  - 9.4 × 10^-14 = 0

x = 3.1 × 10^-7 M

Recall  [BH] = [OH^-] = 3.1 × 10^-7 M

Now;

pOH = - log [OH^-]

pOH = log [3.1 × 10^-7 M]

pOH = 6.5

But;

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 6.5

pH = 7.5

The pH of 455 mg/L of caffeine is 7.5

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Missing parts

Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4.  Calculate the pH of a solution containing a caffeine concentration  of 455 mg/L.

Answer:

= .7567105263

Explanation:

1 atm = 760 mmHg

575.1 mmHg (1 atm/760mmHg) = .7567105263 atm

Answer:

Explanation:

In my opinion the answer should be SO2

Answer:

a should be answer i think.

Answer:

K loses one electron to CI

Explanation:

The lewis electron dot notation shows only the chemical symbol of the element surrounded by dots to represent the valence electrons.

  We have atom of K with one valence electrons

   Cl with 7 valence electrons

For an electrostatic attraction to occur, both particles must be charged. To do this, one of the species must lose an electron, and the other gains it.

This will make both species attain a stable octet;

   Hence, K will lose 1 electron and Cl will gain the electrons.

Answer:

C: K loses one electron to CI

Explanation:

I took the test and got it correct!!

Answer:

1) Increases

2) decreases

3) increases

4) decreases

Explanation:

When the intermolecular forces in a liquid increases, the greater vapour pressure of the liquid decreases accordingly.

Since the vapour pressure is proportional to temperature, as temperature increases, the vapour pressure increases alongside.

As intermolecular forces increases, the boiling point increases accordingly since more energy is required to break intermolecular bonds.

Lastly, the greater the surface area, tell greater the vapour pressure since more liquid surface area is now available.

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

Answer. After cytokinesis is completed at end of meiosis - I two haploid cells are formed.on:

Answer:

C. TWO HAPLOID CELLS ARE FORMED

Explanation:

I TOOK THE EDGUNITY TEST AND I GOT IT CORRECT

Answer:

[tex]p=26683.2Pa[/tex]

Explanation:

Hello,

In this case, since the pressure is computed via:

[tex]p=h*\rho*g[/tex]

Whereas h is the 0.520-m height, [tex]\rho[/tex] is the 13600-kg/m³ density and the g the 9.81-m/s² gravity. Thus, the pressure in Pa is:

[tex]p=0.20m*13,600 \frac{kg}{m^3} *9.81\frac{m}{s^2} \\\\p=26683.2\frac{kg*\frac{m}{s^2} }{m^2} =26683.2\frac{N}{m^2}\\ \\p=26683.2Pa[/tex]

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Answer:

The theoretical yield of CaS is 6.01 g.

Explanation:

The balanced reaction is given as:

[tex]Ca+S\rightarrow CaS[/tex]

The molar mass of Ca and S is 40.08 and 32.065 g/mol respectively.

Number of moles = [tex]\frac{Mass}{Molar Mass}[/tex]

So, 7.19 g of Ca contains [tex](\frac{7.19}{40.08})[/tex] mol of Ca or 0.179 mol of Ca

Also, 2.67 g of S contains [tex](\frac{2.67}{32.065})[/tex] mol of S or 0.0833 mol of S

According to the balanced equation:

1 mol of Ca produces 1 mol of CaS

So, 0.179 mol of Ca produces 0.179 mol of CaS

According to the balanced equation:

1 mol of S produces 1 mol of CaS

So, 0.0833 mol of S produces 0.0833 mol of CaS

As the least number of mol of CaS (product) is produced from S , therefore, S is the limiting reactant.

So, thoretically, 0.0833 mol of CaS is produced.

The molar mass of CaS is 72.143 g/mol.

So, the mass of 0.0833 mol of CaS is [tex](0.0833\times 72.143)[/tex] g or 6.01 g

Hence, the theoretical yield of CaS is 6.01 g.

Answer:

u measure how much power it has

Explanation:

for example u can power a light bulb woth it if u can it was 100eg energy

Answer:

K and ba

Explanation:

Answer: K and Ba

Explanation:

Answer:

molar mass of unknown gas = 1.987 g/mol

Explanation:

First, the number of moles of the unknown gas is found

Using the ideal gas equation: PV = nRT

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

n = PV/RT

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

Molar mass = mass/ number of moles

molar mass = 0.419 g/ 0.2109 mols

molar mass of unknown gas = 1.987 g/mol

The molar mass of unknown gas by using ideal gas equation = 1.987 g/mol.

This equation gives the relation between pressure, volume, temperature as given below:

[tex]PV = nRT[/tex]

P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹

Substitute the above values in the above equation as follows:

n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)

n = 0.2109 moles

[tex]Molar mass = mass/ number of moles[/tex]

Calculate molar mass by using the above equation,

molar mass = 0.419 g/ 0.2109 moles

The molar mass of unknown gas = 1.987 g/mol

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Answer:

29.8

Explanation:

The formula for volume is mass/ density, so 59/1.98. 29.8 is the answer.

Answer:

true

Explanation:

I'm not sure why cause I dont know how to explain but it's TRUE

Answer:

True  

Explanation:

The particles moving faster in a substance the hotter it gets.

Answer:

1 ➡️ Cells

2 ➡️ Arteries

3 ➡️ Veins

4 ➡️ Heart

Explanation:

The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.

The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.

From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood

The electric circuity diagram has the label 1 bulb analogous to cell, label 2 analogous to arteries, label 3 analogous to veins, and label 4 cell analogous to heart.

The electric circuit has been given as the power source and the conducting wires that allows the flow of the current in the circuit.

In the human body, the heart has been transported the oxygenated blood through the arteries to the cell and carried the deoxygenated blood from the cells back to the heart via veins.

In the circuit, the battery has been the source of the power/blood. The current has been carried from the heart to the cell/bulb through the arteries labeled, 2, and transported back to the battery via veins labeled 3.

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Answer:

380

Explanation:

Answer:

3.65 mol O₂

Explanation:

Step 1: RxN

2H₂O₂ → 2H₂O + O₂

Step 2: Define

Given - 7.30 mol H₂O₂

Solve - x mol O₂

Step 3: Stoichiometry

[tex]7.30 \hspace{3} mol \hspace{3} H_2O_2(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} H_2O_2} )[/tex] = 3.65 mol O₂

Answer:

See detailed explanation.

Explanation:

Hello.

In this case, it is firstly necessary to cite that EGS accounts enhanced geothermal systems which are man-made reservoirs, placed where lots of  hot rock is present but there is lack natural permeability, which requires a fluid to be injected into the subsurface to re-open it and therefore creating permeability.

Typically, water has been used for this purpose, but due to the current issue on saving water alternative methods such as supercritical fluids has been being implemented because they have better dynamic properties such as lower viscosities and therefore larger flow velocities, supercritical CO2 is easy and cheap to get as low temperatures are required to turn it in supercritical condition.

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