Help with this please

Answer:

78.125ml

Explanation:

Number of moles in 250ml of 2.5M KCl is (250÷1000)litres×2.5M so we divide these moles by 8M. The answer gotten will be in litres so multiply by 1000 to get it in ml

Answer:

1 would be the answer

protons = atomic # I think

Answer:

hydrogen

Explanation:

hydrogen has the atomic number of 1 which means that hydrogen has one proton and is neutral so it has one electron as well.

Answer:

We can say that we need 171 g barium hydroxide needed to make one model equals solution.

Explanation:

I hope this answered your question.

171 gram of  Ba(OH)2, in grams, is needed to make a 1.00 m aqueous solution.

Mass is the quantity of matter. The mass is calculated by the density divide by the volume.

In 1.00 solution, the mass of Ba(OH)2 is 171 gram.

Thus, the mass of Ba(OH)2 is 171 gram.

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Answer -ˋˏ ༻༺ ˎˊ-

A common use for gravity filtration is for separating anhydrous magnesium sulfate (MgSO4) from an organic solution that it has dried (Figure 1.68b). Anhydrous magnesium sulfate is powdery, and with swirling in an organic solvent creates a fine dispersal of particles like a snow globe.

Answer:

Collision theory states that molecules must collide to react. For most reactions, however, only a small fraction of collisions produce a reaction.

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

[tex]C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M[/tex]

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

[tex]\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g[/tex]

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

[tex]C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M[/tex]

[tex]\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g[/tex]

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

[tex]m = 0.367 g - 0.102 g = 0.265 g[/tex]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

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The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.

Answer:

c

Explanation:

allow plants to absorb water and nutrients from the soil

Answer:

[tex]\boxed {\boxed {\sf 1.6 \times 10^{24} \ atoms \ Si}}[/tex]

Explanation:

We are asked to find how many atoms are in a sample of 75 grams of silicon.

First, we must convert grams to moles using the molar mass. This is the mass of 1 mole of a substance. The molar mass is found on the Periodic Table because it is equal to the atomic mass, but the units are grams per mole instead of atomic mass units.

Look up the molar mass of silicon.

We convert using dimensional analysis, so we must set up a conversion factor.

[tex]\frac {28.085 \ g \ Si}{1 \ mol \ Si}[/tex]

We are converting 75 grams of silicon to moles, so we multiply by this value.

[tex]75 \ g \ Si*\frac {28.085 \ g \ Si}{1 \ mol \ Si}[/tex]

Flip the conversion factor so the units of grams of silicon cancel.

[tex]75 \ g \ Si*\frac {1 \ mol \ Si}{28.085 \ g \ Si}[/tex]

[tex]75 *\frac {1 \ mol \ Si}{28.085 }[/tex]

[tex]\frac {75}{28.085} \ mol \ Si[/tex]

[tex]2.670464661 \ mol \ Si[/tex]

Next, we convert moles to atoms using Avogadro's Number, or 6.022 ×10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are atoms of silicon.

Set up another conversion factor, this time with Avogadro's Number.

[tex]\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}[/tex]

Multiply by the number of moles we found.

[tex]2.670464661 \ mol \ Si*\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}[/tex]

The units of moles of silicon cancel.

[tex]2.670464661 *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 }[/tex]

[tex]2.670464661 * {6.022 \times 10^{23} \ atoms \ Si}[/tex]

[tex]1.60815382 \times 10^{24} \ atoms \ Si[/tex]

The original value of grams (75) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredths place tells us to leave the 6 in the tenths place.

[tex]1.6 \times 10^{24} \ atoms \ Si[/tex]

The molecules of the two gases will have the same kinetic energy at room temperature.

The average kinetic energy of gaseous molecules can be calculated using the formula:

Kinetic Energy = 3/2RT, where R = constant and T = temperature in Kelvin

This means that the kinetic energy of a gaseous molecule is dependent on the temperature of the molecule only.

In other words, the molecular mass of molecules of gases has no bearing on the kinetic energy of each molecule.

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Answer:

3O2→ 2O3

Explanation:

by multiplying 2 by 3 we get 6 on both sides

Answer:

D, two of the above, oxygen and hydrogen.

Explanation:

To solve this problem we can either use orbital stuff or common knowledge. Oxygen and hydrogen are known to form O2 and H2, while sodium forms metallic bonds between the atoms.

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V[/tex]

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V[/tex]

The mole fraction of nitrogen in the mixture is:

[tex]X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33[/tex]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

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Answer:

C6H5S^-

Explanation:

This is because you are removing the hydrogen to make it a conjugate base.

The formula for conjugate base of C6H5SH is C6H5S⁻.

Conjugate base is defined as the substance which is formed when an acid liberate its protons.

The acid donate protons and the acid changes into base.

Conjugate acid is defined as the pair of compound that differs by its protons by gaining protons.

Conjugate acid and base are based in acid base theory.

Thus, the formula for conjugate base of C6H5SH is C6H5S⁻.

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Answer:

At temperatures > 400 °C, CCl4 reacts with H2O over a MgO catalyst to yield HCl and CO2.

Answer:

5.627 × 1019 milliliters

Explanation:

is this what you wanted?

[tex]\boxed{\sf °F=\dfrac{9}{5}°C+32}[/tex]

[tex]\\ \sf\longmapsto °F=\dfrac{9}{5}(197)+32[/tex]

[tex]\\ \sf\longmapsto °F=\dfrac{1773}{5}+32[/tex]

[tex]\\ \sf\longmapsto °F=354.6+32[/tex]

[tex]\\ \sf\longmapsto °F=386.6°F[/tex]

[tex]\rule{200pt}{5pt}[/tex]

[tex]\boxed{\sf K=°C+273}[/tex]

[tex]\\ \sf\longmapsto K=197+273[/tex]

[tex]\\ \sf\longmapsto K=470K[/tex]

Answer:

LICL as an Ionic compounds has covalent ability and they are soluble in polar solvent such as CH3COCH3 but they are insoluble in non-polar solvents. Due to their polarity, CH3COCH3 will decreases the electrostatic forces of attraction thereby resulting in free ions in aqueous solution.

Water is the known as a polar solvent that can dissolve an ionic compounds very easy and as a polar solvent, the arrangement of oxygen and hydrogen atoms in water is in bent shape.

Ionic compounds like LICL that is very polar is soluble in the polar solvent water.

Polar solvents like CH3COCH3 will dissolve polar and ionic solutes because of the attraction of the opposite charges on the solvent and solute particles.

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Explanation:

Answer:

There are 95 atoms in mercury phosphate

Explanation:

You just add the atomic number for both elements

Answer:

B) Ivory

Explanation:

Ivory is a hard, white material from the tusks and teeth of animals, that consists mainly of dentine, one of the physical structures of teeth and tusks.

Platinum, Silver, and Gold are precious metals, meaning they are rare metallic chemical elements of high economic value, shiny, hard, strong with high melting points

Ivory is regarded as the least of the options. Ivory consists of dentine and

are usually found in the tusks and teeth of animals such as elephants. They

however don't have much value as the other metals in the options.

Precious metals  include the following:

They are used for ornamental purposes. They are usually mined and refined

to produce a variety of objects.

Their use are however legal and they are expensive which makes the Ivory

the least out of the options provided.

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Answer:

C. Both depend on how much matter an object has.

Explanation:

weight is the gravitational force acting on an object, so it does vary depending on gravity.

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Answer:

Answer: Change in speed = 11 m/sTime taken = 3.4sHence, acceleration = (11/3.4) m/s² = 3.235 m/s²

Explanation:

(winks and runs off)

Acceleration is 3.235 m/s² when Usain Bolt went from 0.0 to 11.0 m/s in 3.4 seconds.

Acceleration is defined as a rate at which the velocity changes with time with respect to both speed and direction.An object is said to be accelerated if it's speeds up r slows down.

Motion along a circle is always accelerated even when speed is constant it's direction is continuously changing.Acceleration is a vector quantity as it has both magnitude as well as direction.The magnitude of acceleration is described by Newton's second law of motion.

Formula of acceleration is change in speed /time.Substituting the given values in the given formula,(11.0-0.0)/3.4=3.235 m/s².

Thus, the acceleration  is 3.235 m/s² when Usain Bolt went from 0.0 to 11.0 m/s in 3.4 seconds.

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Answer:

We can say that we need 171 g barium hydroxide needed to make one model equals solution.

Explanation:

I hope this answered your question.

Answer:

c>b=a

Explanation:

It is important to note that mass does not affect the average motion/energy per molecule, but temperature does. the higher the temperature the faster the particles are. A has the same temperature as B, so they have the same amount of motion. C is warmer than A and B, so the average motion of the particles in beaker C is the largest

Answer:

Calculate the freezing point of a solution of 3.46 g of a compound,X,in 160 g of benzene. When separate sample of X was vaporised, its density was found 3.27 g/L at 116 c and 773 torr. The freezing point of pure benzen is 5.45 c, and kf is 5 .12 c/m.

Explanation:

First determine molar mass from

P*molar mass = density*R*T

Solve for molar mass. Then

moles = grams/molar mass

Solve for moles

molality = moles/kg solvent

Solve for molality

delta T = Kf*m

Solve for delta T, then convert to freezing point

Explanation:

Leopards have 32 teeth, 4 of which are long, pointed canine teeth. Canine teeth are used to kill prey. Other teeth are for cutting flesh and grinding bone

Answer:

It is the one that has start and finish and a compass on it...I am sorry I can share images for some reason. And underneath the compass, it says 1.0 Mile. It also has a compass in the bottom right-hand corner.

Explanation:

I just took the quiz and got it correct!

The predicted concentration of IO₃⁻ needed to start the precipitation titration is 1.703 × 10⁻⁷ M

The dissociation of zinc (II) iodate can be expressed as:

[tex]\mathbf{Zn (IO_3)_2 \to Zn^{2+} + 2IO_3^-}[/tex]

Given that the solubility product constant Ksp value = 3.9 × 10⁻⁶

For the above dissociation,

[tex]\mathbf{Ksp = [Zn^{2+ }] [IO_3^-]^2}[/tex]

Since [tex]\mathbf{ [Zn^{2+ }] = [Zn(NO_3)_2] = 0.229 \ M}[/tex]

∴

[tex]\mathbf{3.9\times 10^{-8} =(0.229) \times [IO_3^-]^2}[/tex]

[tex]\mathbf{ [IO_3^-]^2 = \dfrac{3.9\times 10^{-8} }{(0.229)}}[/tex]

[tex]\mathbf{ [IO_3^-]^2 = 1.703 \times 10^{-7} \ M}[/tex]

Therefore, we can conclude that the initial predicted concentration of IO₃⁻ needed to start the precipitation titration is 1.703 × 10⁻⁷ M

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Answer:

3

Explanation: