PLEASE HELP

A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature

is increased to 30.0°C. What is the new pressure of the gas in Pa?

(5 Points)

394 kPa

532 Pa

3.94 x 10^5 Pa

Answer:

1.477mole

Explanation:

First, we'll begin by calculating the molar mass of CO2. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the question = 65g

Mole of CO2 =?

Number of mole = Mass/Molar Mass

Mole of CO2 = 65/44

Mole of CO2 = 1.477mole

Therefore, 65g of carbon dioxide (CO2) contains 1.477mole

Answer: [NO2][O2]/[NO] [O3]

Explanation: Kc = [NO2][O2]/[NO] [O3]

The representative particle for copper metal, Cu, refers to the smallest unit which retains the properties of copper – in this case, a single copper atom.

In chemistry, a representative particle refers to the smallest unit of a substance that still retains the properties of that substance. For a pure metal like copper (Cu), the representative particle is an atom. Therefore, the representative particle for copper metal, Cu, is a copper atom. Copper atoms join together in a crystal lattice structure forming the bulk material we see and use in everyday life like wires, coins etc. But the smallest unit of this structure, the unit which represents its basic physical and chemical properties, is a single copper atom.

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Enthalpy change is a measurement of heat energy absorbed or released during a chemical reaction at constant pressure. It reflects the net energy required to break the chemical bonds of reactants versus the energy released when new bonds form in products, determining if the reaction is endothermic or exothermic.

It represents the difference between the enthalpy (heat content) of the reactants and the products. When chemical bonds are broken in reactants, energy is absorbed, and when new bonds form in products, energy is released. If the energy required to break bonds is greater than the energy released when new bonds are formed, the reaction is endothermic and the enthalpy change is positive; conversely, if more energy is released during bond formation than is absorbed during bond breaking, the reaction is exothermic and exhibits a negative enthalpy change.

The enthalpy change can be calculated using the formula ΔH = Σ (bond dissociation energies of reactants) minus Σ (bond dissociation energies of products), which simplifies the overall energy balance of bond breaking and bond formation processes. This understanding allows scientists to predict whether a reaction will absorb or release energy and is crucial in fields such as thermodynamics and chemical engineering.

Answer:

hydrogen

Explanation:

Answer:

Explanation:

Marine debris consists of most anything used and discarded by humans. Winds, rivers and storms can carry this debris to the ocean from far inland, so it is not solely a case of pollution along shorelines. It is also true that oceangoing ships directly contribute to marine debris while they are at sea and not associated with any particular nation.

Answer:

Excess Vinegar

Explanation:

You will have excess vinegar

Answer:

B. A disease wipes out several plant species.

Explanation:

This will be the most detrimental to the environment as it will decrease oxygen levels and many animals will lose their shelter.

- everything else happens every year and is common and normal.

A disease wiping  out several plant species will most likely cause the sustainability of an ecosystem to weaken.

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.

Learn more about ecosystem,here:

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Answer : The temperature of the air in the tire is, 341 K

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

[tex]P\propto T[/tex]

or,

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure = 198 kPa

[tex]P_2[/tex] = final pressure = 225 kPa

[tex]T_1[/tex] = initial temperature = [tex]27^oC=273+27=300K[/tex]

[tex]T_2[/tex] = final temperature = ?

Now put all the given values in the above equation, we get:

[tex]\frac{198kPa}{300K}=\frac{225kPa}{T_2}[/tex]

[tex]T_2=340.9K\approx 341K[/tex]

Therefore, the temperature of the air in the tire is, 341 K

Answer:

0.0144 L

Explanation:

Step 1:

Data obtained from the question. This includes:

Temperature (T) = 298k

Pressure (P) = 1 bar

Time (t) = 1.25 hr

Current (I) = 0.0500 A

Step 2:

Determination of the quantity of electricity (Q) used. This is illustrated below:

Q = it

Time (t) = 1.25 hr = 1.25 x 3600 = 4500 secs

Current (I) = 0.0500 A

Quantity of electricity (Q) =?

Q = it

Q = 0.05 x 4500

Q = 225C

Step 3:

Determination of the number of mole of O2 liberated in the process.

In the electrolytic process, O2 will be liberated according to the equation:

2O^2- + 4e- —> O2

From the above illustration, 4 faraday are needed to liberate 1 mole of O2.

1 faraday = 96500C

Therefore of 4 faraday = 4x96500C = 386000C

From the above equation,

386000C of electricity liberated 1 mole of O2.

Therefore, 225C will liberate = 225/386000 = 5.83x10^-4 mole of O2.

Step 4:

Determination of the volume of the O2 liberated.

Number of mole (n) = 5.83x10^-4 mole

Temperature (T) = 298k

Pressure (P) = 1 bar = 0.987 atm

Gas constant (R) = 0.082atm.L/Kmol

Volume (V) =?

Applying the ideal gas equation:

PV = nRT

The volume of O2 can be obtained as follow:

PV = nRT

0.987 x V = 5.83x10^-4 x298x0.082

Divide both side by 0.987

V = (5.83x10^-4 x298x0.082)/0.987

V = 0.0144 L

The volume of [tex]\(O_2\)[/tex] produced is approximately [tex]0.0145 \ liters[/tex] at [tex]298 \ K[/tex] and [tex]1.00[/tex] bar.

To determine the volume of [tex]\(O_2\)[/tex] gas produced in an electrolytic cell, we can follow these steps:

Calculate the total charge

The total charge [tex]\(Q\)[/tex] passed through the cell is given by the product of current [tex]\(I\)[/tex] and time [tex]\(t\)[/tex]:

[tex]\[ Q = I \cdot t \][/tex]

Given:

[tex]Current, \(I = 0.0500 \, \text{A}\)[/tex]

[tex]Time, \(t = 1.25 \, \text{hr} = 1.25 \times 3600 \, \text{s} = 4500 \, \text{s}\)[/tex]

[tex]\[ Q = 0.0500 \, \text{A} \times 4500 \, \text{s} = 225 \, \text{C} \][/tex]

Determine the amount of [tex]\(O_2\)[/tex] produced

The half-reaction for the production of [tex]\(O_2\)[/tex] in water electrolysis is:

[tex]\[ 2 \, \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \, \text{H}^+ + 4 \, \text{e}^- \][/tex]

From this reaction, we see that [tex]4[/tex] moles of electrons produce [tex]1[/tex] mole of [tex]\(O_2\)[/tex]

The Faraday constant [tex](\(F\))[/tex] is [tex]\(96485 \, \text{C/mol}\)[/tex].

The moles of electrons [tex](\(n_{\text{e}^-}\))[/tex] can be calculated as:

[tex]\[ n_{\text{e}^-} = \frac{Q}{F} = \frac{225 \, \text{C}}{96485 \, \text{C/mol}} = 0.00233 \, \text{mol} \][/tex]

From the stoichiometry of the half-reaction, [tex]4[/tex] moles of electrons produce [tex]1[/tex] mole of [tex]\(O_2\)[/tex]:

[tex]\[ n_{\text{O}_2} = \frac{n_{\text{e}^-}}{4} = \frac{0.00233 \, \text{mol}}{4} = 0.000583 \, \text{mol} \][/tex]

Calculate the volume of [tex]\(O_2\)[/tex] gas

Use the ideal gas law to find the volume of [tex]\(O_2\)[/tex] gas at [tex]298\ K[/tex] and [tex]1.00[/tex] bar:

[tex]\[ PV = nRT \][/tex]

Given:

Pressure, \[tex](P = 1.00 \, \text{bar} = 1.00 \times 10^5 \, \text{Pa}\)[/tex]

Temperature, [tex]\(T = 298 \, \text{K}\)[/tex]

Gas constant, [tex]\(R = 8.314 \, \text{J/(mol K)}\)[/tex]

Moles of [tex]\(O_2\), \(n = 0.000583 \, \text{mol}\)[/tex]

[tex]\[ V = \frac{nRT}{P} = \frac{0.000583 \, \text{mol} \times 8.314 \, \text{J/(mol K)} \times 298 \, \text{K}}{1.00 \times 10^5 \, \text{Pa}} \][/tex]

[tex]\[ V = \frac{1.447 \, \text{J}}{1.00 \times 10^5 \, \text{Pa}} = 1.447 \times 10^{-5} \, \text{m}^3 \][/tex]

[tex]\[ V = 0.01447 \, \text{L} \][/tex]

Answer:

Explanation:

SO3 (g) + NO (g) U SO2 (g) + NO2 (g)

occurs under these conditions. Calculate the value of the equilibrium constant, Kc, for the above reaction.

SO3 (g) + NO (g) U SO2 (g) + NO2 (g)

Initial (M) 2.00 2.00 0 0

Change (M) −x −x +x +x

Equil (M) 2.00 − x 2.00 − x x x

2 2 c 3

2

c 2

[SO ][NO ]

[SO ][NO]

(2.00 )

=

= −

K

x K

x

Since the problem asks you to solve for Kc, it must indicate in the problem what the value of x is. The concentration of

NO at equilibrium is given to be 1.30 M. In the table above, we have the concentration of NO set equal to 2.00 − x.

2.00 − x = 1.30

x = 0.70

Substituting back into the equilibrium constant expression:

2c 2 2c 2

(2.00 )

(0.70)

(2.00 0.70)

= − = −

x KxK

Kc = 0.290

Answer:

It is not possible, since the total weight of all the reagents involved in the reaction is equal to the total weight of the product, since it is not possible to create or destroy atoms, thus, it is not possible to change the weight of the product. It is equal to the weight of the reagent.

Explanation:

Answer:

Approximately [tex]\rm 31.5\; g[/tex].

Explanation:

The mass of a solution can be divided into two parts:

In this particular [tex]\rm KOH[/tex] solution in water,

The number [tex]35\%[/tex] here likely refers to the concentration of [tex]\rm KOH[/tex] in this solution. That's ratio between the mass of the solute ([tex]\rm KOH[/tex]) and the mass of the whole solution (mass of solute plus mass of solvent.) That is:

[tex]\displaystyle \frac{m(\text{KOH})}{m(\text{solution})} = 35\% = 0.35[/tex].

Hence, [tex]m(\mathrm{KOH}) = 0.35\, m(\text{solution})[/tex].

However, since the solution contains only the solute and the solvent, [tex]m(\text{solution}) = m(\text{solute}) + m(\text{solvent})[/tex].

For this solution in particular,

[tex]\begin{aligned}&m(\text{solution})\\&= m(\text{solute}) + m(\text{solvent}) \\ &= m(\text{KOH}) + m(\text{water})\end{aligned}[/tex].

As a result,

[tex]\begin{aligned}&m(\mathrm{KOH})\\ &= 0.35\, m(\text{solution}) \\&= 0.35\, (m(\mathrm{KOH}) + m(\text{water}))\\&= 0.35\, m(\mathrm{KOH}) + 0.35 \, m(\text{water})\end{aligned}[/tex].

Subtract [tex]0.35\, m(\mathrm{KOH})[/tex] from both sides of the equation:

[tex](1 - 0.35)\, m(\mathrm{KOH}) = 0.35\, m(\text{water})[/tex].

[tex]\begin{aligned} &m(\mathrm{KOH}) \\ &= \left(\frac{0.35}{1 - 0.35}\right)\cdot m(\text{water}) \\ &= \frac{0.35}{0.65} \times 58.5\; \text{g} = 31.5 \; \text{g}\end{aligned}[/tex].

Note, that for this calculation, there's nothing special about this [tex]35\%[/tex] solution of [tex]\mathrm{KOH}[/tex] in water. In general,

[tex]\displaystyle m(\text{solute}) = \left(\frac{\%\text{concentration}}{100\% - \%\text{concentration}}\right)\cdot m(\text{solvent})[/tex].

Answer:

32g

Explanation:

We have to remember that for percent (w/w) concentration we usually write;

Percent concentration= mass of solute/mass of solution ×100

Since mass of solute= 14.7 g and percent concentration = 32.2%

Then

Mass of solution= mass of solute × 100/ percent concentration

Mass of solution= 14.7 ×100/32.2

Mass of solution= 46.7 g

Since mass of solution = mass of solute + mass of solvent

Mass of solute= 14.7 g

Mass of solution = 46.7g

Mass of solvent = 46.7g -14.7g = 32g

Answer:

After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.

Explanation:

Answer:

73.4% is the percent yield

Explanation:

2KClO₃ →  2KCl  + 3O₂

This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.

We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃

In the theoretical yield of the reaction we say:

2 moles of potassium chlorate can produce 3 moles of oxygen

Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂

The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g

But, we have produced 115 g. Let's determine the percent yield of reaction

Percent yield = (Produced yield/Theoretical yield) . 100

(115g / 156.6g) . 100 = 73.4 %

Final answer:

The recoil speed of the gun is found using the conservation of momentum. The calculation shows that the gun's recoil speed is approximately 2.4 ms-1. So the correct option is A.

Explanation:

The question posed is related to the concept of conservation of momentum in physics. When a bullet is fired from a gun, the bullet moves forward and the gun is pushed back due to recoil, both objects conserving the system's total momentum. To find the speed of recoil of the gun, you can use the formula:

[tex]m_{bullet}v_{bullet} = m_{gun}v_{gun}[/tex]

Where:

Thus, solving for [tex]v_{gun}[/tex] :

[tex]v_{gun} = m_{bullet}v_{bullet} / m_{gun}[/tex]

[tex]v_{gun}[/tex]= (0.01 kg * 120 m/s) / 0.49 kg = 2.45 m/s

The speed of the recoil of the gun is therefore approximately 2.4 ms-1, which matches answer choice A.

Answer: heating the canister

Answer:

heating the canister

Explanation:

Answer:

All helium atoms have two protons, and no other elements have atoms with two protons. In the case of helium, the atomic number is 2. The atomic number of an element is usually written in front of and slightly below the element's symbol, like in the Figure below for helium.

Answer:

The standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s) is -0.744 V

Explanation:

Here we have

1. Cr³⁺  + e − ⟶ Cr²⁺ E⁰₁ = − 0.407 V

2. Cr²⁺ + 2 e − ⟶ Cr ( s ) E⁰₂  = − 0.913 V

To solve the question, we convert, the E⁰ values to ΔG as follows

ΔG₁ = n·F·E⁰₁ and ΔG₂ = n·F·E⁰₂

Where:

F = Faraday's constant in calories

n = Number of e⁻

ΔG₁ = Gibbs free energy for the first reaction

ΔG₂ = Gibbs free energy for the second half reaction

E⁰₁  = Reduction potential for the first half reaction

E⁰₂ = Reduction potential for the second half reaction

∴ ΔG₁ = 1 × F × − 0.407 V

ΔG₂ = 2 × F  × − 0.913 V

ΔG₁  + ΔG₂  = F × -2.233 V which gives

ΔG = n × F × ΔE⁰ = F × -2.233 V  

Where n = total number of electrons ⇒ 1·e⁻ + 2·e⁻ = 3·e⁻ = 3 electrons

We have, 3 × F × ΔE⁰ = F × -2.233 V

Which gives ΔE⁰ = -2.233 V /3 = -0.744 V.

The standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s) is found by the summation of the standard reduction potentials of the half-reactions involved, giving -1.32V.

To calculate the standard reduction potential for the reduction half-reaction of Cr(III) to Cr(s), we first need to understand that a redox reaction is a sum of an oxidation half-reaction and a reduction half-reaction. In the given half-reactions, the first is for Cr3+ being reduced to Cr2+ and the second one is for Cr2+ being reduced to Cr(s).

From the given standard reduction potentials, the 1st reaction has E°1 = -0.407 V and the 2nd reaction has E°2 = -0.913 V. Therefore, to get from Cr3+ to Cr(s), we add both these half-reactions together, which also means we add their potentials together. The sum gives us the potential for the entire reaction, which is E°total = E°1 + E°2 = (-0.407V) + (-0.913V) = -1.32V.

Thus, the standard reduction potential for the reduction half-reaction of Cr(III) to Cr(s) is -1.32V.

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Answer:

Succinate oxidation to fumarate The following reactions transform succinate to regenerate oxalacetate. The first of these reactions is carried out by an oxidation catalyzed by succinate dehydrogenase. The hydrogen acceptor is FAD, since the free energy change is insufficient to allow NAD to interact. The final product is fumarate.

Explanation:

The condensation reaction of GDP + Pi and the hydrolysis of Succinyl-CoA involve the H2O necessary to balance the equation.

Answer:

the answer is option 3.

Answer: 1.1g of copper

Explanation: We begin by writing the balanced reaction equation.

So we have : Cu(s) + 2AgNO3 (aq) ------> Cu(NO3)2 (aq) + 2Ag (s)

Molar mass of AgNO3 = 107.87 + 14.01 + 3(16.0) = 169.88g/mol, for 2 moles we then have 2 x 169.88 =339.76g.

Atomic mass of Copper = 64g approximately

From the equation the following deductions can be made:

339.76g of AgNO3 reacts with 64g of copper

5.65g of AgNO3 would react with 64/ 339.76 x 5.65 =1.062 approx 1.1g of copper.

Answer:

we need 1.06g of Cu to react with 5.65g of AgNO₃

Explanation:

C u +2 AgNO₃ = Cu(NO₃ )₂ + 2 Ag

the mole ratio between Copper and silver nitrate is 1:2  meaning we need  2 moles of silver nitrate for every mole of copper that takes part in the reaction.

the molar mass of Cu  is 63.55

the molar mass of AgNO₃ is 107.8682+ 14 + (16x3) = 107.8682+ 14 + 48=  169.87682≈169.88g

for every 63.55 g of Cu reaction with  2(169.88g of AgNO₃)= 339.736g of AgNO₃

if x grams of Cu react with 5.65 grams of AgNO₃

63.55g= 339.736g

x = 5.65g

cross multiply

339.736x = 5.65 x 63.55 = 359.0575

x = 359.0575/339.736 = 1.05687210069≈1.06

we need 1.06g of Cu to react with 5.65g of AgNO₃

Answer:

a) 13.2 moles [tex]2H_{2}O[/tex]

b) 79.33 grams of [tex]2H_{2}O[/tex]

Explanation:

First, we'll need to balance the equation

[tex]H_{2(g)} + O_{2(g)}[/tex] → [tex]H_{2}O_{(g)}[/tex]

There are 2 (O) on the left and only one on the right, so we'll add a 2 coefficient to the right.

[tex]H_{2(g)} + O_{2(g)}[/tex] → [tex]2H_{2}O_{(g)}[/tex]

Now there are 4 (H) on the right and only 2 on the left, so we'll add a 2 coefficient to the ([tex]H_{2}[/tex]) on the left.

[tex]2H_{2(g)} + O_{2(g)}[/tex] → [tex]2H_{2}O_{(g)}[/tex]

The equation is now balanced.

a) This can be solved with a simple mole ratio.

4.6 moles [tex]O_{2}[/tex] × [tex]\frac{2 moles H_{2}O}{1 mole O_{2}}[/tex] = 13.2 moles [tex]2H_{2}O[/tex]

b) This problem is solved the same way!

2.2 moles [tex]H_{2}[/tex] × [tex]\frac{2 moles H_{2}O}{2 moles H_{2}}[/tex] = 2.2 moles [tex]2H_{2}O[/tex]

However, this problem wants the mass of [tex]2H_{2}O[/tex], not the moles.

The molecular weight of [tex]2H_{2}O[/tex] is the weight of 4 (H) molecules and 2 (O) molecules (found on the periodic table). So,

4(1.008) + 2(15.999) = 36.03 g/mol

2.2 moles [tex]2H_{2}O[/tex] × [tex]\frac{36.03 g}{1 mol}[/tex] = 79.33 grams of [tex]2H_{2}O[/tex]

Answer:

128g of Li, will react in this reaction

Explanation:

Before to start working, we need the reaction:

N₂ and Li react, in order to produce Li₃N (lithium nitride)

N₂ + 6Li → 2Li₃N

1 mol of nitrogen reacts with 6 moles of lithium

We convert the mass of N₂ to moles → 86.1 g . 1 mol/ 28g = 3.075 moles

1 mol of N₂ reacts with 6 mol of Li

Therefore, 3.075 moles of N₂ will react with 18.4 moles of Li

We conver the moles to mass → 18.4 mol . 6.94g / 1mol = 128 g

The two solutions AgNO3 and K2CrO4 react to give a precipitate along with an aqueous solution of KNO3. Hence, after the reaction is complete, KNO3 remains in the solution.

When the two aqueous solutions AgNO3 and K2CrO4 react, they produce solid silver chromate Ag2CrO4 as a precipitate and leave potassium nitrate KNO3 in the solution. The reaction is balanced as 2AgNO3 (aqueous) + K2CrO4 (aqueous) -> Ag2CrO4 (s) + 2KNO3 (aqueous). As the reaction proceeds, silver ions (Ag+) and chromate ions (CrO42-) combine to form the precipitate, leaving potassium ions (K+) and nitrate ions (NO3-) in the solution. Hence, after the reaction, the solution consists of KNO3 because it remains aqueous, with 1.00 M K+ and 1.00 M NO3-.

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Answer : The final pressure of the gas is, [tex]2.47\times 10^3mmHg[/tex]

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]

or,

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure = 545 mmHg

[tex]P_2[/tex] = final pressure = ?

[tex]V_1[/tex] = initial volume = 775 mL

[tex]V_2[/tex] = second volume = 171 mL

Now put all the given values in the above equation, we get:

[tex]545mmHg\times 775mL=P_2\times 171mL[/tex]

[tex]P_2=2470.03mmHg=2.47\times 10^3mmHg[/tex]

Therefore, the final pressure of the gas is, [tex]2.47\times 10^3mmHg[/tex]

Answer:

Their pH is less than 7

Explanation:

Both of these compounds pHs are greater than 7.

Answer:

A) Their pH is less than 7

Explanation:

Their pH is less than 7 is not correct. Their pH is greater than seven, but not equal to seven.