a boy capable of swimming 2.1m/a in still water is swimming in a river with a 1.8 m/a current. At what angle must he swim in order to end up directly opposite his starting point?

Answer:

4.22

Explanation:

According to the question, the pOH of the solution is 9.78. You may recall that pOH is the hydroxide concentration of a solution.

Also pOH = -log[OH^-]. Hence the pOH is obtained from the hydroxide ion concentration.

Finally, pH + pOH =14

Hence;

pH = 14-pOH

pH= 14-9.78 = 4.22

pH= 4.22

Answer:

The correct answer is 2.2 mL.

Explanation:

Given:

Average: 2.9 mL

SD: 0.71 mL

We can define a 1 SD range in which the value of volume (in mL) will be comprised:

Volume (mL) = Average ± SD = (2.9 ± 0.7) mL

Maximum value= Average + SD= 2.9 + 0.7 mL = 3.6 mL

Minimum value= Average - SD = 2.9 - 0.7 mL = 2.2 mL

Thus, the minimum value within a 1 SD range of the average is 2.2 mL

The minimum value within 1 SD is 2.19 mL

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x\ is\ raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]

Given that μ = 2.9 mL, σ = 0.71 mL; hence:

The minimum value within 1 SD range = μ ± σ = 2.9 ± 0.71 = (2.19, 3.61)

Therefore the minimum value within 1 SD is 2.19 mL

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Answer:

7.4 × 10⁻⁹ M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷

Concentration of lithium ion: 0.0020 M

Step 2: Write the reaction for the solution of Li₃PO₄

Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

Step 3: Calculate the phosphate concentration required for a precipitate to occur

The solubility product constant is:

Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³

[PO₄³⁻] = 7.4 × 10⁻⁹ M

Answer:

(a )people separate mixtures in order to ger a specific substance that they need.

Answer:

I need great answers

Explanation:

please rate my answer as great

Answer:

Entropy change of ice changing to water at 0°C is equal to 57.1 J/K

Explanation:

When a substance undergoes a phase change, it occurs at constant temperature.

The entropy change Δs, is given by the formula below;

Δs = q/T

where q is the quantity of heat absorbed or evolved in Joules and T is temperature in Kelvin at which the phase change occur

From the given data, T = 0°C = 273.15 K, q = 15.6 KJ = 15600 J

Δs = 15600 J / 273.15 K

Δs = 57.111 J/K

Therefore, entropy change of ice changing to water at 0°C is equal to 57.1 J/K

The entropy change of ice changing to water will be "57.1 J/K".

The shift in what seems like a thermodynamic system's condition of confusion is caused by the transformation of heat as well as enthalpy towards activity. Entropy seems to be greater mostly in a network with a high quantity or measure of chaos.

According to the question,

Temperature, T = 0°C or,

                          = 273.15 K

Heat, q = 15.6 KJ or,

            = 15600 J

We know the formula,

Entropy change, Δs = [tex]\frac{q}{T}[/tex]

By substituting the values, we get

                                 = [tex]\frac{15600}{273.15}[/tex]

                                 = 57.11 J/K

Thus the above answer is correct.    

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Answer:

Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)

Explanation:

The ionic equation shows the actual reaction that took place. It excludes the spectator ions. Spectator ions are ions that do not really participate in the reaction even though they are present in the system.

For the reaction between iron and copper II nitrate, the molecular reaction equation is;

Fe(s) + Cu(NO3)2(aq)----> Fe(NO3)2(aq) +Cu(s)

Ionically;

Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)

Answer:

hello the molecules are missing from your question below are the Generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]

answer : It can be determined  that both generic molecules are polar

It can be determined that both generic molecules have similar molecular shape

They have different Geometry

They differ in bond angles as well

Explanation:

The two generic molecules : [tex]ABE_{3}[/tex] and [tex]AB_{3} E[/tex]

comparing(similarities) these two generic molecules

It can be determined  that both generic molecules are polar

It can be determined that both generic molecules have similar molecular shape

differences between the generic molecules

They have different Geometry

They differ in bond angles as well

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater

mass of freshwater = density * volume

1 cm³ = 1 mL

mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g

mass of freshwater + cup = 734.265 + 25 = 759.265 g

Therefore,  mass of equal volume of seawater = 759.265 g

Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)

1 liter = 1000 cm³ = 1000 mL;

Density of seawater = mass / volume

Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L

Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L

mass of 1 Liter seawater = 1033.01 g

mass of 1 Liter freshwater = 999 g

mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g

Therefore, amount of salt in 1 L seawater = 34 g

Answer:

The empirical formula of the compound is NO2.

Explanation:

The following data were obtained from the question:

Mass of compound = 30.5 g

Mass of nitrogen (N) = 9.29 g

Empirical formula of compound =?

Next, we shall determine the mass of oxygen in the compound. This can be obtained as follow:

Mass of compound = 30.5 g

Mass of nitrogen (N) = 9.29 g

Mass of oxygen (O) =?

Mass of O = mass of compound – mass of N.

Mass of O = 30.5 – 9.29

Mass of O = 21.21 g

Finally, we shall determine the empirical formula of the compound as follow:

Mass of nitrogen (N) = 9.29 g

Mass of oxygen (O) = 21.21 g

Divide by their molar mass.

N = 9.29 / 14 = 0.664

O = 21.21 / 16 = 1.326

Divide by the smallest

N = 0.664/ 0.664 = 1

O = 1.326/ 0.664 = 2

Therefore the empirical formula of the compound is NO2.

Answer:

2 Na(s) + C(s) + 3/2 O₂(g) → Na₂CO₃(s)

Explanation:

The molar enthalpy of formation of a chemical is defined as the change in enthalpy during the formation of 1 mole of the substance from its constituent elements (Constituent elements are pure elements you have in the periodic table)

For Na₂CO₃ constituent elements are Na(s), C(s) and O₂(g) and the chemical equation that represents the molar enthalpy is:

Answer:

Explanation:

Before proceeding to answering the questions, let's try and understand each term.

Gravitational energy can be defined as the potential energy an object has when placed in a high position (compared to been placed in a low position). Examples include; a knife been placed at the top of the cupboard, a ball at the top of a hill.

Nuclear energy can be defined as the energy released or consumed as a result of splitting or joining together of the nuclei (plural of nucleus) of atom(s). The process of splitting an atomic nucleus is called nuclear fission while the process of joining two nuclei of atoms is called nuclear fusion.

Radiant energy can be defined as the energy of electromagnetic wave. One of the properties of electromagnetic waves is that they travel through space. Examples include X-rays and light

Stored mechanical energy can be defined as the potential energy an object has as a result of the application of force. Examples include; a stretched rubber band, a compressed spring

Answer:

[Cu²⁺] = 2.01x10⁻²⁶

Explanation:

The equilibrium of Cu(CN)₄²⁻ is:

Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻

And Kf is defined as:

Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴

As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:

[Cu²⁺] = 0

[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M

[Cu(CN)₄²⁻] = 2.2x10⁻³

Some [Cu²⁺] will be formed and equilibrium concentrations will be:

[Cu²⁺] = X

[CN⁻] = 0.3212M + 4X

[Cu(CN)₄²⁻] = 2.2x10⁻³ - X

Where X is reaction coordinate

Replacing in Kf equation:

1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴

1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0

Solving for X:

X = 2.01x10⁻²⁶

As

[Cu²⁺] = X

Answer:

B

Explanation:

Increasing distance between the objects factor will cause the gravitational force between two objects to decrease. Therefore, option B is correct.

The gravitational force grows in proportion to the size of the masses . The gravitational force weakens rapidly as the distance between masses grows. Unless at least one of the objects has a lot of mass, detecting gravitational force is extremely difficult.

Gravity is affected by object size and distance between objects. Mass is a unit of measurement for the amount of matter in an object.

The force of gravity is proportional to the masses of the two objects and inversely proportional to the square of the distance between them. This means that the force of gravity increases with mass but decreases as the distance between objects increases.

Thus, option B is correct.

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#SPJ6

Answer:

The correct answer is -341.2 kJ per mole.

Explanation:

The reaction given is:  

CO (g) + Cl₂ (g) ⇔ COCl₂ (g)

Kp = 5.62 × 10³⁵

T = 25 °C or 298 K

The formula for calculating ΔG is,  

ΔG° = -RTlnKp

ΔG° = -8.314 × 298 ln (5.62 × 10^35)

ΔG° = -203.9 kJ/mol

ΔG° = ∑nΔG°f (products) -∑nΔG°f (reactants)

ΔG° = ΔG°f (COCl₂ (g)) - [ΔG°f (CO(g)) + ΔG°f (Cl₂(g))]

ΔG°f (COCl₂ (g)) = ΔG° + [ΔG°f (CO (g)) + ΔG°f (Cl₂(g))]

ΔG°f (COCl₂ (g)) = -203.9 + (-137.28 + 0.00)

ΔG°f (COCl₂ (g)) = -341.2 kJ/mol

The standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol

The given equation for the chemical reaction is

CO(g) + Cl2(g) ⇌ COCl2(g)

At the temperature of 25°C = (273 + 25) K, the equilibrium constant [tex]\mathbf{K_p = 5.62\times 10^{35}}[/tex]

Consider the expression for the relationship between [tex]\mathbf{\Delta G^o}[/tex] and [tex]\mathbf{K_p }[/tex] for the equilibrium reaction can be expressed as:

[tex]\mathbf{\Delta G^o = - RT In K_p}[/tex]

where;

∴

[tex]\mathbf{\Delta G^o = - (8.314 \times 10^{-3}\ kJ/K.mol \times 298 \ K) \times In (5.62 \times 10^{35} )}[/tex]

[tex]\mathbf{\Delta G^o = -2.477572\ K \times 82.31680992}[/tex]

[tex]\mathbf{\Delta G^o = 203.95 \ kJ}[/tex]

Thus, the standard free energy for the reaction is 203.95 kJ/mol

For a given reaction, the standard Gibbs free energy can be calculated by using the formula:

[tex]\mathbf{\Delta G^o_{rxn} = \sum n \Delta G^o_f (products) - \sum m \Delta G^o_f (reactants) }[/tex]

[tex]\mathbf{\Delta G^o_{rxn} =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(\Delta G^o_{f} (CO)_{(g)} + \Delta G^o_{f} (Cl)_{2(g)} ) \Big ) \Big ] }[/tex]

replacing the values of and solving for COCl2 at standard free energy of formation of substances, we have:

[tex]\mathbf{-203.95 \ kJ/mol =\Big [\Delta G^o_{f} (COCl_{2(g)} ) -\Big(-137.3 kJ/mol + 0 \ kJ/mol\Big ) \Big ] }[/tex]

Collecting like terms, we have:

[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -203.95 \ kJ/mol -137.3 kJ/mol }[/tex]

[tex]\mathbf{\Delta G^o_{f} (COCl_{2(g)} ) = -341.25 \ kJ/mol }[/tex]

Therefore, we can conclude that the standard Gibbs free energy [tex]\mathbf{\Delta G^o_f}[/tex] for COCl2 at 25°C is -341.25 kJ/mol

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Answer:

[tex]0.10 \text{ L$^2$mol$^{-2}$s$^{-1}$}[/tex]

Explanation:

The general formula for a rate law is

[tex]\text{rate} = k\text{[A]}^m \text{[B]}^{n}[/tex]

With your numbers, the rate law becomes

1.2 mol·L⁻¹s⁻¹ = k(2 mol·L⁻¹)²(3 mol·L⁻¹)¹ = k × 4 mol²L⁻² × 3 mol·L⁻¹

= 12k mol³L⁻³

[tex]\\ k = \dfrac{\text{1.2 mol $\cdot$ L$^{-1}$s$^{-1}$} }{12\text{ mol$^{3}$L}^{-3}} = \mathbf{0.10} \textbf{ L$\mathbf{^2}$mol$^{\mathbf{-2}}$s$^{\mathbf{-1}}$}[/tex]

Answer:

The correct answer is CaO > LiBr > KI.

Explanation:

Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.  

With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.  

The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.  

Arranging the chemical compounds in order of decreasing magnitude of lattice energy, we have:

c. CaO.

a. LiBr

b. KI

Lattice energy can be defined as a measure of the energy required to dissociate one (1) mole of an ionic compound into its constituent anions and cations, in the gaseous state.

Hence, it is typically used to measure the bond strength of ionic compounds.

Generally, lattice energy is inversely proportional to the size of the ions and directly proportional to their electric charges.

Lithium bromide (LiBr) comprises the following ions:

Potassium iodide (KI) comprises the following ions:

Calcium oxide (CaO) comprises the following ions:

From the above, we can deduce that there is an increase in the charge possessed by the ionic chemical compounds and as such this would result in an increase in the lattice energy.

In order of decreasing magnitude of lattice energy, the chemical compounds are arranged as:

I. CaO.

II. KI.

III. LiBr.

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Answer:

(CH3)3N

Explanation:

Hydrogen bonding can be called a type of intracellular force of the attraction. It is the force that occur between molecules. It is the bonding between the molecules and of hydrogen and electronegative items in the covalent bond. This is called the hydrogen donor. An electro-negative hydrogen atoms may be a hydrogen bonded. It is also called a hydrogen acceptor.

Thus in (CH3)3N, the hydrogen atoms becomes bonded with carbon. Carbon is not electronegative atoms. Thus it does not play as donor. Nitrogen is electronegative and play as hydrogen acceptor. But there is no presence of hydrogen acceptor. Thus there is no molecules that exhibit hydrogen molecules bonding.

[tex]\bold {(CH_3)_3N}[/tex] does not have hydrogen bonding because of the absence of electronegativity difference.

Hydrogen bond:

It is an inter-molecular bond. It is due to the difference in electronegativities of constituent atoms. This creates dipole in the atoms so, atoms start to attract each other.

In [tex]\bold {(CH_3)_3N}[/tex], the hydrogen atoms are bonded with carbon. The difference between the electronegativities Carbon and hydrogen is very less.

Therefore, [tex]\bold {(CH_3)_3N}[/tex] does not have hydrogen bonding because of the absence of electronegativity difference.

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Answer:

222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.

Explanation:

Being:

the molar mass of chromium (II) nitrate, Cr(NO₃)₂ is:

Cr(NO₃)₂ = 52 g/mole + 2* (14 g/mole + 3* 16 g/mole)= 176 g/mole

So: if 176 grams are present in 1 mole of the compound, 5.08 grams in how many moles of the compound will be present?

[tex]amount of moles=\frac{5.08 grams* 1 mole}{176 grams}[/tex]

amount of moles=0.0289 moles

Molarity (M) is the number of moles of solute that are dissolved in a given volume. It is then calculated by dividing the moles of the solute by the volume of the solution:

[tex]molarity (M)=\frac{number of moles of solute}{volume}[/tex]

Molarity is expressed in [tex]\frac{moles}{liter}[/tex]

So in this case:

Replacing:

[tex]0.130 M=0.130 \frac{moles}{liter} =\frac{0.0289 moles}{volume}[/tex]

Solving:

[tex]volume=\frac{0.0289 moles}{0.130 \frac{moles}{liter} }[/tex]

volume=0.2223 liters

Being 1 L= 1,000 mL:

volume=0.222 liters= 222.3 mL

222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.

Answer:

[tex]\huge\boxed{Option \ 2}[/tex]

Explanation:

A reaction is spontaneous at all temperatures by the following combinations:

=> A negative enthalpy change ( [tex]\triangle H < 0[/tex] )

=> A positive entropy change ( [tex]\triangle S > 0[/tex] )

See the attached file for more better understanding!

from Gibbs Equation, [tex] \Delta G = \Delta H - T\Delta S [/tex]

reaction is spontaneous if $\Delta G$ is negative.

so, first option is not valid at high temperature, ($-h+ts$)

second, is always a spontaneous reaction, ($-h-ts$)

third, is never spontaneous ($+h+ts$)

4th is similar to second, spontaneous at higher temperatures ($+h-ts$)

Answer:

length

Explanation:

edge 2020

hope this helps!

Answer:

A.) Density

Explanation:

Correct on edge.

Answer:

b.) Put on protective gloves

Answer:

2020 Put on protective gloves.

Explanation:

Answer:

[tex]\Delta G=-97.14kJ[/tex]

Explanation:

Hello,

In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

[tex]\Delta G=-RTln(K)[/tex]

Hence, we compute it as required:

[tex]\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol[/tex]

And for 2.37 moles of hydrogen bromide, we obtain:

[tex]\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ[/tex]

Best regards.

Answer:

See explanation

Explanation:

First voltaic cell;

Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)

Anode;

Zinc

Cathode;

Copper

Oxidation half equation;

Zn(s)------> Zn^2+(aq) + 2e

Reduction half equation;

Cu^2+(aq) +2e -----> Cu(s)

Overall; Zn(s) + Cu^2+(aq) -----> Zn^2+(aq) + Cu(s)

E°cell = 0.34 -(-0.76) =1.1 V

Second voltaic cell;

Zn(s)|Zn^2+(aq)||Fe^2+(aq)|Fe(s)

Anode;

Zinc

Cathode;

Iron

Oxidation half equation;

Zn(s)------> Zn^2+(aq) + 2e

Reduction half equation;

Fe^2+(aq) +2e -----> Fe(s)

Overall; Zn(s) + Fe^2+(aq) -----> Zn^2+(aq) + Fe(s)

E°cell = (-0.44) -(-0.76) = 0.32 V

Third voltaic cell;

Fe(s)|Fe^2+(aq)||Cu^2+(aq)|Cu(s)

Anode;

Iron

Cathode;

Copper

Oxidation half equation;

Fe(s)------> Fe^2+(aq) + 2e

Reduction half equation;

Cu^2+(aq) +2e -----> Cu(s)

Overall; Fe(s) + Cu^2+(aq) -----> Fe^2+(aq) + Cu(s)

E°cell = 0.34 -(-0.44) = 0.78 V

Fourth voltaic cell

Cu(s)|Cu^2+(aq)||I2(aq)|C(s)|I^-(aq)

Anode;

Copper

Cathode;

Graphite rod

Oxidation half equation;

Cu(s)------> Cu^2+(aq) + 2e

Reduction half equation;

I2(aq) +2e -----> 2I^-(aq)

Overall; Cu(s) + I2(aq) -----> Cu^2+(aq) + 2I^-(aq)

E°cell = 0.54 -0.34 = 0.20 V

Answer:

- photography     (captures images that can be shared and

compared by scientists)

- space telescope    (allows access to images taken from outside

Earth's atmosphere)

- radio telescope     (detects electromagnetic frequencies outside

the visible spectrum that reaches Earth)

- optical telescope    (obtains a magnified and clear view of a part

of the sky to observe celestial objects)

Explanation:

Photography is used to capture still images based on the principle that some compounds react in the presence of optical energy.

Space telescope is a type of observatory telescope positioned in outer space to observe distant planets, galaxies and other astronomical objects. Space telescopes reduces the interference from ultraviolet frequencies, X-rays and gamma rays; as well as light pollution which ground-based observatories encounter.

A radio telescope is a specialized antenna and radio receiver used to receive radio waves from astronomical radio sources in the sky. Radio telescope is used to study radio frequencies  emitted by astronomical objects, that fall outside the visible light spectrum.

An optical telescope is used to gather and focuses light, from a far distant object. Optical telescope is used within the visible light spectrum of the electromagnetic spectrum, to create a magnified image, for direct view, or to make a photograph, or to collect data through electronic image sensors.

Answer:

- photography     (captures images that can be shared and

compared by scientists)

- space telescope    (allows access to images taken from outside

Earth's atmosphere)

- radio telescope     (detects electromagnetic frequencies outside

the visible spectrum that reaches Earth)

- optical telescope    (obtains a magnified and clear view of a part

of the sky to observe celestial objects)

Explanation:

Answer:

[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]

Explanation:

A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.

[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]

Answer:

The amount of moles of gaseous boron trifluoride, BF₃, contained in a 4.3421 L bulb at 787.9 K if the pressure is 1,218 atm is 0.082 moles

Explanation:

An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

Replacing:

1.218 atm* 4.3421 L= n*0.082 [tex]\frac{atm*L}{mol*K}[/tex] *787.9 K

Solving:

[tex]n=\frac{1.218 atm* 4.3421 L}{0.082 \frac{atm*L}{mol*K}*787.9 K}[/tex]

n= 0.082 moles

The amount of moles of gaseous boron trifluoride, BF₃, contained in a 4.3421 L bulb at 787.9 K if the pressure is 1,218 atm is 0.082 moles