1. A mixture of 3 gases (A, B, and C) has a total pressure of 240kPa. If the partial pressure of
Gas A is 107 kPa and the partial pressure of Gas B is 55 kPa, what is the partial pressure of
Gas C?

Answer:

Weight,mass, volume, density

Answer:

Mass, volume, weight, and density.

Explanation:

Alright, let's take it step by step!

(a) When you mix water with different temperatures, the final temperature is like a weighted average. Imagine you have `v` liters of water at temperature `T` and `u` liters of water at temperature `S`. The amount of thermal energy in the first batch is `v*T` and in the second batch it's `u*S`. When you combine them, the total thermal energy is `v*T + u*S`. Since the total volume is now `v + u`, the average energy per liter (which is the final temperature) is `(v*T + u*S) / (v + u)`.

In equation form:

Final Temperature, F = (v*T + u*S) / (v + u).

(b) Now let's move to the changing volumes and temperatures. Let `v(t)` be the volume at time `t`, and `T(t)` the temperature at time `t`. Let's say that in `Gt` seconds, `Gu` gallons of water are added at temperature `S`. We’ll assume that 1 gallon is the same as 1 liter for simplicity, though in reality they are slightly different.

The new volume after `Gt` seconds is `v(t) + Gu`, and the total thermal energy is `v(t)*T(t) + Gu*S`. The new average temperature is:

T(t+Gt) = (v(t)*T(t) + Gu*S) / (v(t) + Gu).

Now, T(t+Gt) - T(t) = [(v(t)*T(t) + Gu*S) / (v(t) + Gu)] - T(t).

Now, let's think about water pouring at a constant rate. Let's use the limit definition of the derivative. Instead of `Gu` gallons in `Gt` seconds, let's say a tiny amount of water `du` is added in a tiny amount of time `dt`. So, `du/dt` is the rate at which water is poured into the urn.

Using the limit definition:

dT/dt = lim (dt -> 0) [(v(t)*T(t) + du*S) / (v(t) + du) - T(t)] / dt

     = [(v(t)*T(t) + du*S) / (v(t) + du) - T(t)]' (derivative with respect to t)

     = [v'(t)*T(t) + v(t)*T'(t) + du/dt*S - v'(t)*T(t) - v(t)*T'(t)] / (v(t) + du) (using product rule)

     = (du/dt*S) / (v(t) + du).

As dt approaches 0, du becomes very small, and thus we can ignore it in comparison to v(t), so:

dT/dt ≈ (du/dt*S) / v(t).

This is the rate of change of temperature with respect to time, in terms of the rate at which water is poured, the temperature at which it is poured, and the volume of water already in the urn.

To answer this question we have to use the rule of dilutions:

Where V1 is the initial volume, C1 is the initial concentration, V2 is the final volume and C2 is the final concentration. In this case we need to find V1.

Solve the equation for V1 and replace C1 for 4.00M, V2 for 665mL and C2 for 0.350M

It means that the volume required is 58.2mL.

Answer:

300 mL

Explanation:

the unit formula of calcium phosphate is Ca3(PO4)2

molar mass of Ca3(PO4)2 = (3×40 + 2×31 + 8×16) g/mol = 310 g/mol

n = m/M = 35 g/(310 g/mol)

c = n/V

V = n/c = [35 g/(310 g/mol)]/0.375 mol/L

V = 0.30 L = 300 mL

The portion of energy from the Sun that reaches the Earth​ is known as

electromagnetic spectrum.

Energy transfer from objects in space such as the Sun usually reach the

earth through the process known as radiation. The energy is referred to as

electromagnetic energy.

Components of the electromagnetic spectrum are:

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The correct answer is:

(c). Bacterial hydrolysis of ingested labeled urea producing labeled bicarbonate, which is absorbed into the blood and exhaled as 14CO₂ or 13CO₂.

The Helicobacter pylori (H. pylori) infection can be found using the urea breath test. The bacteria H. pylori has the ability to colonize the stomach and result in a number of gastrointestinal disorders. The urease enzyme produced by bacteria is what causes the urea breath test to detect its presence.

The patient takes a small amount of urea that has been labeled with a radioactive or non-radioactive carbon isotope (often 13C or 14C) as part of the test. The urease enzyme produced by H. pylori will hydrolyze the urea in the stomach if the bacteria is there, creating labeled bicarbonate (CO₂) as a byproduct.

The blood then carries this labeled bicarbonate to the lungs, where it is exhaled as either 14CO₂ or 13CO₂. Breath samples taken from the patient are examined for the presence of the tagged carbon dioxide. The presence of an H. pylori infection is indicated if the tagged carbon dioxide is found.

As a result, the procedure involved in the urea breath test for H. pylori detection is accurately described by option c. Options a, b, and d do not adequately describe how the urea breath test is conducted.

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Answer:

That is def true

Explanation:

Answer:

ok so basicalkly he lloked at how the subdiction zone formed and where previus conintal drift occured

Explanation:

3) orbitals can hold two electrons provided they are spinning in the same direction

We define an orbital in chemistry as a region in space where there is a high probability of finding an electron.

Each orbital holds a maximum of two electrons which spin in opposite direction according to Pauli exclusion principle.

According to this principle no two electrons can have the same set of quantum numbers. As a result, the spin quantum numbers of two electrons in the same orbital must differ hence they spin in opposite directions.

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Explanation:

Ions form when atoms gain or lose electrons. This is so that they form a full outer shell of electrons. When an atom gains electrons it becomes a negative ion, because electrons are negatively charged. For example, all halogens (group 7 or 17) form negative ions as they gain an electron forming a 1- charge. When an atom loses electrons it becomes a positive ion, as it is losing some negative charge from the electrons. This would be for example, alkali metals (group 1) which lose an electron to form a positive ion with a 1+ charge, (ALL metals form positive ions).

The boiling point of the sucrose solution would be 3.68 °C compared to the boiling point of pure water.

The boiling point of a solution depends on the concentration of solute particles in the solution, which affects the colligative properties of the solution, such as boiling point elevation.

To calculate the boiling point elevation, we can use the following formula;

ΔTb = Kbm

where ΔTb is the boiling point elevation, Kb is the molal boiling point constant for water (which is a constant value), and bm is the molality of the solution, which is the amount of solute (in moles) per kilogram of solvent.

First, let's calculate the molality of the solution;

moles of sucrose = mass of sucrose / molar mass of sucrose

moles of sucrose = 739 g / 342.3 g/mol (molar mass of sucrose)

Next, let's convert the mass of water to kilograms:

mass of water = 0.300 kg

Now, we can calculate the molality of the solution;

bm = moles of sucrose / mass of water

Plugging in the values;

moles of sucrose = 739 g / 342.3 g/mol ≈ 2.158 mol

mass of water = 0.300 kg

bm = 2.158 mol / 0.300 kg ≈ 7.193 mol/kg

Next, we need to determine the molal boiling point constant (Kb) for water. The molal boiling point constant for water is approximately 0.512 °C kg/mol.

Finally, we can calculate the boiling point elevation;

ΔTb = Kb × bm

ΔTb = 0.512 °C kg/mol × 7.193 mol/kg

≈ 3.68 °C

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Answer:

there are 4 sulfur atom.

Answer:

overall I think one answer

Option (d) is correct. As a polar molecule, water cannot easily dissociate inorganic compounds.

Ionic compound is defined as a chemical compound composed of ions held together by electrostatic forces termed ionic bonding. The ionic compounds are neutral.  It consists of positively charged ions called cations and negatively charged ions called anions. polarity is explained as a separation of electric charge leading to a molecule or its chemical groups having an electric dipole moment with a negatively charged end and a positively charged end. These polar molecules must contain one or more polar bonds due to a difference in electronegativity between the bonded atoms. water cannot easily dissociate inorganic compound due to polarity.

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The complete question is,

Which response provides the best explanation as to why ionic compounds easily dissociate in water?

(a) nonpolar organic molecules such as fats and waxes dissolve well in water.

(b)  the polarity of water allows it to easily dissociate most covalently bound compounds.

(c) the polarity of water easily breaks the charges between the oppositely charged ions in the compound.

(d) as a polar molecule, water cannot easily dissociate inorganic compounds.

To find the empirical formula from a molecular formula, you should divide the Subscripts by the Some integer in order to get the smallest whole number.

Part-A

The empirical formula of \(Al_{2}Br_{6}\)is \(AlBr_{3}\), we get it by dividing both subscript (i.e. 2&6) by 2.

Part-B

The empirical formula of C8H10 is

Simplest ratio of CH = 8:10 = 4:5

Hence, empirical formula =C4H5

Part-C

The empirical formula of C4H8O2 is C2H4O, we get it by diving all subscripts \(\frac{4}{2} : \frac{8}{2} :\frac{2}{2}=2:4:1\) by 2.

Part- D

molecular formula= P4O10

simplest ratio of

\(\frac{P}{O}= \frac{2}{5} \\ empirical formula = p_{2}O_{5}\)

Part -E

The empirical formula of C6H9CL2 is C3H2CL,we get it by diving all subscripts\(\frac{6}{2}:\frac{4}{2}:\frac{2}{1} =3:2:1\) by 2

Part-F

The empirical formula of B3N3H6is BNH2, we get it by diving all subscripts \(\frac{3}{3} :\frac{3}{3} :\frac{6}{3} = 1:2:3\) by 2.

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To determine whether each statement about rain or acid rain is true or false. Please find the evaluation of each statement below:

1. There is more acid rain in the Western vs. Eastern United States.
False. There is more acid rain in the Eastern United States due to higher concentrations of sulfur oxides and nitrogen oxides from industrial activities and power plants.

2. Unpolluted rain water is acidic and approximately 3.6 on the pH scale.
False. Unpolluted rainwater is slightly acidic with a pH of around 5.6. This is due to the natural presence of carbon dioxide in the atmosphere, which dissolves into water and forms carbonic acid.

3. Sulfur oxides and nitrogen oxides in the troposphere cause acid rain.

True. Sulfur oxides (SOx) and nitrogen oxides (NOx) are released from various sources like industrial processes, power plants, and vehicle emissions. These gases react with water, oxygen, and other substances in the troposphere to form sulfuric acid and nitric acid, which then fall to the earth as acid rain.

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The mass of the cell wall, assuming the cell is spherical and the wall is a thin spherical shell, is approximately 0.91 milligrams.

To calculate the mass of the cell wall, we first need to determine the volume of the wall.

The given diameter of the cell is 0.00 μm, which means the radius (r) of the cell is half of that, so r = 0.00/2 = 0.00 μm = 0.00 nm.Now, we need to find the volume of the cell wall, which can be approximated as a thin spherical shell. The volume of a thin spherical shell can be calculated using the formula:

V = 4/3 * π * (r_outer^3 - r_inner^3)

Since the cell is spherical, the inner radius of the shell is the same as the radius of the cell (r), and the outer radius of the shell is the sum of the radius of the cell (r) and the thickness of the wall (60.0 nm). Thus, the outer radius (r_outer) of the shell is:

r_outer = r + thickness = 0.00 + 60.0 = 60.0 nm

Substituting the values into the formula, we have:

V = 4/3 * π * (60.0^3 - 0.00^3)

= 4/3 * π * (216,000 nm^3)

= 288,000 π nm^3

Next, we need to calculate the mass of the cell wall using the density of pure water. The density (ρ) is given as 1000 kg/m^3, which is equivalent to 1000 kg/1,000,000,000 nm^3 since 1 m = 1,000,000,000 nm. Thus, the mass (m) of the cell wall is:

m = ρ * V

= 1000 kg/1,000,000,000 nm^3 * 288,000 π nm^3

= 0.000288 π kg

Now, we can calculate the mass of the cell wall by substituting the value of π (pi) as 3.14159:

m = 0.000288 * 3.14159 kg

= 0.000905 kg

≈ 0.91 mg

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Answer:

D. 30 g

Explanation:

Conversion of mass ensures that though the sugar may appear gone the particles remain and will only add onto the original liquid

Answer:

30 grams

Explanation:

The water level raises 30 grams because that was how much Erica added into the water. good luck and I hope this helped

Answer:

\(1 \: ml = \frac{1}{1000} l \\ 7.65 \times {10}^{5} \: ml = \frac{7.65 \times {10}^{5} }{1000} \\ = 765 \: l\)

Answer:

Metal

Non-metal

Explanation:

Ionic compounds are said to be formed when metals donate electrons to non metals.

The metals become positively charged while the nonmetals become negatively charged.

CaF2 is composed of Ca^2+ and F^-. Calcium is a metal of group 2 in the periodic table while F is a nonmetal of group 17 in the periodic table.

Hence CaF2 is composed of a metal and a nonmetal.

Answer: Here are some examples of atoms that behave similarly to chlorine in terms of electron affinity:

Fluorine (F) has the highest electron affinity of any element, so it is more electronegative than chlorine. However, fluorine and chlorine are both halogens, which means that they have similar chemical properties.

Bromine (Br) is also a halogen, and it has a very similar electron affinity to chlorine. In fact, bromine is often used as a substitute for chlorine in organic chemistry.

Iodine (I) is the third halogen, and it has a slightly lower electron affinity than chlorine. However, iodine is still a very electronegative element, and it behaves similarly to chlorine in many chemical reactions.

Nitrogen (N) is not a halogen, but it has a relatively high electron affinity. This is because nitrogen has a small atomic radius, which means that its valence electrons are held more loosely than the valence electrons of larger atoms.

Oxygen (O) is also not a halogen, but it has a relatively high electron affinity. This is because oxygen has a small atomic radius and it also has two unpaired valence electrons.

Explanation: Fluorine has the highest electron affinity, followed by chlorine, bromine, and iodine.

Nitrogen and oxygen also have high electron affinities because they have small atomic radii and unpaired valence electrons.

Atoms with high electron affinity are more likely to attract electrons, which means they are more electronegative.

After 800 years, approximately 59.02% of the original amount of radium-226 will remain. The rest would have decayed into other elements or isotopes.

The half-life of radium-226 is 1620 years, which means that after 1620 years, half of the original amount will decay. Therefore, after 800 years, we can calculate the percentage of radium-226 that remains using the formula:

Remaining percentage = (1/2)^(800/1620) x 100

Plugging in the numbers, we get:

Remaining percentage = (1/2)^(0.493827) x 100

Remaining percentage = 59.02%

So, after 800 years, approximately 59.02% of the original amount of radium-226 will remain. The rest would have decayed into other elements or isotopes.

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The molarity of the solution is approximately 5.0 M.

To calculate the molarity (M) of a solution, we need to divide the moles of solute by the volume of the solution in liters.

First, we need to convert the mass of NaOH to moles. The molar mass of NaOH is 22.99 g/mol for Na, 16.00 g/mol for O, and 1.01 g/mol for H, giving us a total molar mass of 39.99 g/mol for NaOH.

Moles of NaOH = 120.0 g / 39.99 g/mol

= 3.00 mol

Next, we divide the moles of NaOH by the volume of the solution in liters:

Molarity (M) = moles of solute / volume of solution in liters

M = 3.00 mol / 9.60 L

= 0.3125 M

Rounding to the appropriate number of significant figures, the molarity of the solution is approximately 5.0 M.

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Answer:

11.3 g

Explanation:

7.895 + 3.4 = 11.295

When rounded to correct number of significant figures --> 11.3

There are 3 significant figures in 11.3

Answer:

11.3 g when rounded

Answer:

d i took the test

Explanation:

Answer:

2RbNO2 + BaCO3 → Rb2CO3 + Ba(NO2)2

Explanation:

The balanced reaction equation is shown below;

2RbNO2 + BaCO3 → Rb2CO3 + Ba(NO2)2

This reaction is possible because the reduction potential of Rb is -2.98V while that of Ba is –2.92 V. Hence Rb can displace Ba from its salt solution.

The equation is balanced since the number of atoms of each element on the left and right hand sides of the reaction equation are equal.

The products for the reaction of Rubidium nitrite and Barium carbonate has been Barium nitrite and Rubidium carbonate.

Rubidium nitrite has been the binary ionic compound found to be acidic in nature. While barium carbonate has been the basic salt with the potency to be poisonous to humans.

The reaction between an acid and a base has been the neutralization reaction with the formation of salt and water. The reaction between rubidium nitrite and barium carbonate will be:

\(\rm 2\;RbNO_2\;+\;BaCO_3\;\rightarrow\;Rb_2CO_3\;+\;Ba(NO_2)_2\)

The reaction has been the displacement reaction with the tendency to displacement.

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Option (d), P=ATC=MR=MC, accurately represents the long-run equilibrium condition for perfect competition, reflecting the balance between price and cost for firms operating in a competitive market.

The long-run equilibrium condition for perfect competition is that price (P) is equal to average total cost (ATC), which is also equal to marginal cost (MC), and marginal revenue (MR).

Option (d), P=ATC=MR=MC, best represents the long-run equilibrium condition for perfect competition. In perfect competition, firms operate at the minimum point of their average total cost curve, where price equals both average total cost and marginal cost. This condition ensures that firms are earning zero economic profit and are producing at an efficient level.

In the long run, if firms are earning economic profit, new firms will enter the market, increasing competition and driving prices down. Conversely, if firms are experiencing losses, some firms may exit the market, reducing competition and causing prices to rise. This process continues until firms reach a state where price equals average total cost, marginal cost, and marginal revenue, ensuring a long-run equilibrium.

Therefore, option (d), P=ATC=MR=MC, accurately represents the long-run equilibrium condition for perfect competition, reflecting the balance between price and cost for firms operating in a competitive market.

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The heat of reaction (ΔH) for the given reaction is 180.4 kJ/mol. To calculate the heat of reaction (ΔH) for the given reaction:

CCl₄(g) + H₂O(g) -> CHCl₃(g) + HCl(g)

You would need the standard enthalpies of formation for each compound involved in the reaction. The standard enthalpy of formation (ΔHf) is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Here are the standard enthalpies of formation for the compounds involved:

ΔHf[CCl₄(g)] = -135.5 kJ/mol

ΔHf[H₂O(g)] = -241.8 kJ/mol

ΔHf[CHCl₃(g)] = -104.7 kJ/mol

ΔHf[HCl(g)] = -92.3 kJ/mol

To calculate ΔH for the reaction, you need to sum up the enthalpies of formation of the products and subtract the sum of the enthalpies of formation of the reactants:

ΔH = ΣΔHf(products) - ΣΔHf(reactants)

ΔH = [ΔHf[CHCl₃(g)] + ΔHf[HCl(g)]] - [ΔHf[CCl₄(g)] + ΔHf[H₂O(g)]]

ΔH = [(-104.7 kJ/mol) + (-92.3 kJ/mol)] - [(-135.5 kJ/mol) + (-241.8 kJ/mol)]

ΔH = -196.9 kJ/mol - (-377.3 kJ/mol)

ΔH = 180.4 kJ/mol

Therefore, the heat of reaction (ΔH) for the given reaction is 180.4 kJ/mol.

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To solve this problem, let's use the definition for molarity:

Replacing the values of the problem:

Now, to find the mass, we multiply by the molecular weight of NaCl. (Which is about 58.44g/mol)

The answer is approximately 22.2g of NaCl