Help please I really need it. ​

Answer:

In thinking about the overall health of the macroeconomy, it is useful to consider three primary goals: economic growth, full employment (or low unemployment), and stable prices (or low inflation).

To draw a Bohr Rutherford diagram for carbon-12, we would have the symbol C in the center, with 2 electrons in the first energy level and 4 electrons in the second energy level, arranged in pairs. Each energy level would be represented by a circle around the nucleus, with the appropriate number of electrons placed in the circles.

A Bohr Rutherford diagram is a visual representation of the electron arrangement in an atom. Carbon-12 has 6 protons and 6 neutrons, which means it has 6 electrons as well since it is a neutral atom. In a Bohr Rutherford diagram, the nucleus is represented by the symbol for the element and the protons and neutrons are shown as small circles inside the symbol. The electrons are represented as dots or circles around the symbol, in the order of increasing energy levels.

In the case of carbon-12, the first energy level can hold a maximum of 2 electrons, and the second energy level can hold a maximum of 8 electrons. Therefore, the first energy level will have 2 electrons, and the remaining 4 electrons will go in the second energy level.

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Answer:

See explanation

Explanation:

A titration involves the addition of a titrant to an analyte solution. It is a method of volumetric analysis.

When a particular volume of titrant is added, the colour changes to signal the end point of the reaction.

The point at which the colour changes is called the equivalence point. This is the point at which the amount of titrant added is just enough to completely neutralize the analyte solution.

Hence the volume NaOH that needs to be added to the beaker containing HCl to cause a colour change is the volume of NaOH that is just enough to completely neutralize the HCl solution.

Research more Respiration

Answer:

wheres the map? I can't see it even tho im using my eye glasses

The volume of hydrogen gas produced in the reaction of 73 grams of zinc and 73 grams of hydrochloric acid (under normal conditions) is approximately 22.4 liters.

To calculate the volume of hydrogen gas produced in the reaction of zinc and hydrochloric acid, we need to use the principles of stoichiometry and the ideal gas law.

First, let's write the balanced chemical equation for the reaction between zinc (Zn) and hydrochloric acid (HCl):

Zn + 2HCl →\(ZnCl_2\)+ H2

From the equation, we can see that one mole of zinc reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas. To determine the number of moles of zinc and hydrochloric acid, we need to convert the given masses into moles.

The molar mass of zinc (Zn) is approximately 65.38 g/mol, so 73 grams of zinc is equal to:

73 g Zn * (1 mol Zn / 65.38 g Zn) ≈ 1.116 mol Zn

Similarly, the molar mass of hydrochloric acid (HCl) is approximately 36.46 g/mol, so 73 grams of HCl is equal to:

73 g HCl * (1 mol HCl / 36.46 g HCl) ≈ 2.002 mol HCl

According to the balanced equation, the reaction produces one mole of hydrogen gas for every two moles of hydrochloric acid. Therefore, since we have 2.002 moles of HCl, we expect to produce half that amount, or approximately 1.001 moles of hydrogen gas.

To calculate the volume of hydrogen gas, we can use the ideal gas law, which states:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. In this case, we assume the reaction is conducted under normal conditions, which means a pressure of 1 atmosphere and a temperature of 273.15 Kelvin.

Rearranging the equation to solve for V, we have:

V = nRT / P

Substituting the values, we get:

V = (1.001 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm) ≈ 22.4 L

Therefore, the volume of hydrogen gas produced in the reaction is approximately 22.4 liters.

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The metal X has an approximate molar mass of 42.36 g/mol and the metal is most likely calcium.

The molar mass of the metal carbonate XCO₃ and identify the metal X, we need to calculate the number of moles of XCO₃ and CO₂ using the given masses and molar masses.

The molar mass of CO₂ (carbon dioxide) is 12.011 g/mol (for carbon) + 2 * 15.999 g/mol (for oxygen) = 44.01 g/mol.

The number of moles of CO₂ can be calculated using the formula:

moles of CO₂ = mass of CO₂ / molar mass of CO₂

moles of CO₂ = 0.855 g / 44.01 g/mol

moles of CO₂ ≈ 0.01944 mol

Since the reaction stoichiometry is 1:1 between XCO₃ and CO₂, the number of moles of XCO₃ is also approximately 0.01944 mol.

molar mass of XCO₃ = mass of XCO₃ / moles of XCO₃

molar mass of XCO₃ = 2.012 g / 0.01944 mol

molar mass of XCO₃ ≈ 103.38 g/mol

The molar mass of XCO₃ is approximately 103.38 g/mol.

To determine the metal X:

molar mass of X = molar mass of XCO3 - molar mass of CO3

molar mass of X = 103.38 g/mol - (12.011 g/mol + 3 * 15.999 g/mol)

molar mass of X ≈ 42.36 g/mol

Metal X is most likely Calcium that has a molar mass of 40 g/mol

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When 0.20 of nicotinic acid is dissolved in 350 ml of water, the ph is 3.05. calculate the Ka value Ka=1.41ₓ10⁻⁵ and percent ionization is 14.1%

One of the water-soluble B vitamins is niacin, sometimes referred to as vitamin B3. Nicotinic acid (pyridine-3-carboxylic acid), nicotinamide (niacinamide or pyridine-3-carboxamide), and similar compounds, such as nicotinamide riboside, are collectively referred to as "niacin."

pH = 3.05

pH=-log[H⁺]

H⁺=10⁻³·⁰⁵

H⁺=0.00089125M

Moles of nicotinic acid in number =2.0ₓ10⁻²

Water volume = 350 mL = 0.0350 L

percent ionization=1.41/100

                               14.1%

A potent basic substance is nicotinic acid (based on its pKa). From microbes to people, all living things have nicotinic acid.

Molarity =Moles/Volume in Liters

             =2.0ₓ10⁻²/0.0350

Molarity = 0.057214M

Nicotinic acid dissocates as follows:

HA⇒[H⁺]+[A⁻]

[H⁺]=0.00089125M

[A⁻]=0.057214M

At equilibrium, [HA] = 0.05714 - 0.00089125 = 0.05624875 M

Ka=[H⁺][A⁻]/[HA]

Ka=(ₓ0.00089125M)(0.057214M)/0.05624875 M

Ka=1.41ₓ10⁻⁵

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When 2.0 moles of KCIO3 decompose, 2.0 moles of O2 will form.

The balanced chemical equation shows that 2 moles of KCIO3 decompose to produce 3 moles of O2. Therefore, we can use the stoichiometric ratio from the balanced equation to determine the number of moles of O2 formed when 2.0 moles of KCIO3 decompose.

According to the stoichiometry, 2 moles of KCIO3 produce 3 moles of O2. Therefore, we can set up a proportion:

(2 moles KCIO3 / 2 moles O2) = (2.0 moles KCIO3 / x moles O2),

where x represents the unknown number of moles of O2 formed.

Simplifying the equation:

(2 moles KCIO3 / 2 moles O2) = (2.0 moles KCIO3 / x moles O2),

1 = (2.0 moles KCIO3 / x moles O2),

Cross-multiplying:

x moles O2 = (2.0 moles KCIO3 / 1),

x moles O2 = 2.0 moles KCIO3.

Therefore, when 2.0 moles of KCIO3 decompose, 2.0 moles of O2 will form.

It is important to note that this calculation assumes complete and ideal conditions, where the reaction proceeds with 100% efficiency. In reality, the actual yield of O2 may be lower due to various factors such as side reactions or incomplete decomposition. To determine the actual yield, additional information or experimental data would be required.

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The number of lithium, Li atoms in 4.5 moles of lithium, Li is 2.7×10²⁴ atoms

1 mole of Li = 6.02×10²³ atoms

From Avogadro's hypothesis given above,

1 mole of Li = 6.02×10²³ atoms

Therefore,

4.5 moles of Li = 4.5 × 6.02×10²³

4.5 moles of Li = 2.7×10²⁴ atoms

Thus, we can conclude that 2.7×10²⁴ atoms are present in 4.5 moles of lithium.

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Answer:

90.99 or 91.0

Explanation:

Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.

Answer: The volume of oxygen gas is 91.4 L.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\(\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\) ......(1)

Given mass of ethanol = 38.5 g

Molar mass of ethanol = 46 g/mol

Plugging values in equation 1:

\(\text{Moles of ethanol}=\frac{38.5g}{46g/mol}=0.840 mol\)

The given chemical equation follows:

\(CH_3CH_2OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)\)

By stoichiometry of the reaction:

If 1 mole of ethanol produces 3 moles of water

So, 0.840 moles of ethanol will produce = \(\frac{3}{1}\times 0.840=2.52mol\) of water

The ideal gas equation is given as:

\(PV=nRT\) .......(2)

where

P = pressure = 1.75 atm

V = volume of oxygen gas = ?

n = number of moles= 2.52 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = temperature of the tank = \(500^oC=[500+273]K=773K\)

Putting values in equation 2, we get:

\(1.75 atm\times V=2.52mol\times 0.0821L.atm/mol.K\times 773K\\\\V=\frac{2.52\times 0.0821\times 773}{1.75}=91.4L\)

Hence, the volume of oxygen gas is 91.4 L.

Answer:

the community needs to work alongside with government and religious organisations. the communities work on both profile and non profile work incase of human rights violations. they also force the government to take necessary action

In redox reactions, an oxidation half reaction shows the oxusetoon occurring while the reduction half reaction shows the reduction occurring in the overall reaction.

Redox reactions are reactions in which oxidation and reduction reactions occur together to the same extent.

A redox reaction can be split into two half oxidation and reduction reactions.

The half reactions of the given reactions are shown below:

Overall reaction;

\(A. Mg(s) + CoBr_2(aq) ⟶ MgBr_2(aq) + Co(s) \\ \)

Oxidation half reaction:

\(Mg (s) ---> Mg ^{2+}(aq)\)

Reduction half reaction:

\(Co^{2+} (aq) ---> Co (s)\)

B. Overall reaction

\(Cu(s) + 2AuCl(aq) ⟶ CuCl_2(aq) + 2Au(s) \\ \)

Oxidation half reaction:

\(Cu(s) ---> Cu^{2+}(aq)\)

Reduction half reaction:

\(2Au^{+}(aq) ---> 2 Au (s)\)

Therefore, redox equations can be spilt into oxidation and reduction half reactions.

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Thickness of the fail in millimeter is 0.051mm

Thickness is the quality or state of being thick

Here given data is

aluminum(Al) =  24cm by 31cm

Mass =10.35g

Density = 2.70gcm³

We have to find the thickness of the fail in millimeter = ?

So the formula is

thickness = Volume/surface area

So first we have to calculate volume

Volume = mass/density

Volume = 10.35g/ 2.70gcm³

Volume = 3.83

Then the area of foil is

Area = 24cm×31cm

Area = 744cm²

Thickness = volume/area

Thickness = 3.83/744cm²

Thickness = 0.0051cm = 0.051mm

Thickness of the fail in millimeter is 0.051mm

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The balanced chemical equation is 2AlBr₃ + 3K₂SO₄ → 6KBr + Al₂(SO₄)₃

When we look closely at the given reaction which is

AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃

Aluminuim is not balance, potassium is not balanced, bromine is also not balanced as well as oxygen and sulphur.

To do that, let's pick the reactant side

2AlBr₃ + 3K₂SO₄

Adding two moles to the aluminuim bromide will give us 2 atoms of aluminuim and 6 atoms of bromine while 3 moles to potassium sulphate will give us 6 atoms of potassium, 3 atoms of sulphur and 12 atoms of oxygen.

To ensure that law of conservation of matter is maintained here, our product should be

6KBr + Al₂(SO₄)₃

This shows that there's no loss of matter.

The overall balanced chemical equation is

2AlBr₃ + 3K₂SO₄ → 6KBr + Al₂(SO₄)₃

At the end of this, we can see that

2 moles of aluminuim at the reactant is reflected at the product

6 moles of bromine at the reactant is reflected at the product

6 moles of potassium at the reactant is reflected at the product

3 moles of sulphur at the reactant is reflected at the product

12 moles of oxygen at the reactant is reflected at the product

From the explanation above, the balanced equation is

2AlBr₃ + 3K₂SO₄ → 6KBr + Al₂(SO₄)₃

Answer:

Fronts

Explanation:

For example, there are hot and cold fronts which cause the air to become warmer or cooler in a specific region!

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Answer:

the answer is A. cooking the eggs

Answer:

A. Cooking the eggs

Explanation:

Cooking the eggs is a chemical change because it's a permanent change. There's no way to get the eggs back in its original form as an egg. Breaking the egg, stirring the egg, or even putting spices on the eggs are all examples of physical changes because they don't change the eggs' chemical composition. They only change its appearance.

Answer: True

Explanation:

The new volume of the gas is 56.6 liters when the temperature is raised to 325 K and the pressure is lowered to 1.2 atm.

PV = nRT

Where R is the ideal gas constant. Since the number of moles is constant in this problem, we can simplify the ideal gas law to:

P1V1/T1 = P2V2/T2

Where the subscripts 1 and 2 refer to the initial and final states of the gas, respectively.

We can now plug in the given values for the initial state of the gas:

P1 = 2.3 atm

V1 = 25 L

T1 = 299 K

And the given values for the final state of the gas:

P2 = 1.2 atm

T2 = 325 K

We can then solve for V2:

P1V1/T1 = P2V2/T2

(2.3 atm)(25 L)/(299 K) = (1.2 atm)V2/(325 K)

V2 = (2.3 atm)(25 L)(325 K)/(1.2 atm)(299 K)

V2 = 56.6 L (rounded to three significant figures)

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The answer is: The electronic configuration of Nitrogen is \(1s^22s^22p^3\).

Electronic configuration: The electronic configuration is defined as the distribution of electrons of an atom in the atomic or molecular orbitals and is written using the labels for the subshell.

How to decide which orbital is filled first?

(Shown in image)

         \(1s^22s^22p^3\)

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Nitrogen's complete electron configuration is 12s2s22p3.

The shorthand electron configuration for noble gases is [He] 2s22p3. Nitrogen has an atomic number of 7. The nitrogen atoms' nucleus contain this many protons. An atom that is neutral has an equal number of protons and electrons. Thus, the ground state electron configuration will consist of 7 electrons in the suitable s and p orbitals (state of lowest energy). For nitrogen, the entire electron configuration is 1s22s22p. Scientists may easily express and explain how the electrons are organized around the nitrogen atom's nucleus by using the configuration notation for nitrogen (N). As a result, it is simpler to comprehend and forecast how atoms will cooperate to form chemical bonds.

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Volume:-

\(\\ \tt\longmapsto 0.333(0.444)(0.555)=0.08cm^3\)

So

\(\\ \tt\longmapsto Density=\dfrac{Mass}{Volume}\)

\(\\ \tt\longmapsto Density=\dfrac{2.22}{0.08}=27.75g/cm^3\)

The cations can be identified by using various test, that are explained in the below section, Cations such as Na+, Ca2+, Cr3+, Fe3+, Al3+ and many other cation's test are explained below.

Na+ produces yellow flames in flame test. So it confirms the presence of Na+ ion.

If we add NaOH in sample and it produces vapor and turns red litmus, then NH4+ is present. But Here there is no change in litmus. So NH4+ ion CANNOT be present.

The addition of HCl does NOT produces white precipitate indicates that Ag+ CANNOT be present there. When we add HCl in Ag+ , a white precipitate of AgCl is formed.

The addition of buffered NH/NH3 to a portion of the solution produces a precipitate. Cr3+ , Fe3+ , Al3+ produces precipitate in addition of NH/NH3 buffer. So Cr3+ , Fe3+ , Al3+ CAN be present there but further test can confirm it.

Ammonium oxalate produces white precipitate in presence of calcium ion. So Ca2+ must be present there.

The addition of a strong base will precipitate Mg2+ and Ni2+ ion if they are present and Zn2+ forms soluble compound . But here they do not formed precipitate hence Mg2+ and Ni2+ CANNOT be Present there but Zn2+ CAN be present there , we need further tests to confirm Zn2+ ion.

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Answer: the bottom one is wave the one above is electromagnetic and the one above is transverse  

Explanation:

There are approximately \(2.76 * 10^{24\) ammonia molecules in the given sample.

To calculate the number of ammonia molecules in the sample, we need to use Avogadro's number and the molar mass of ammonia.

The molar mass of ammonia \((NH_3)\) can be calculated by adding up the atomic masses of nitrogen (N) and hydrogen (H):

Molar mass of \(NH_3\) = (1 x atomic mass of N) + (3 x atomic mass of H)

              = (1 x 14.01 g/mol) + (3 x 1.01 g/mol)

              = 14.01 g/mol + 3.03 g/mol

              = 17.04 g/mol

Now, we can calculate the number of moles of ammonia in the sample using the formula:

Number of moles = Mass of the sample / Molar mass

Number of moles = 78.25 g / 17.04 g/mol

              ≈ 4.5865 mol (rounded to four decimal places)

Finally, we can use Avogadro's number, which represents the number of particles (atoms, molecules, etc.) in one mole of a substance. Avogadro's number is approximately \(6.022 * 10^{23\) particles/mol.

Number of ammonia molecules = Number of moles x Avogadro's number

Number of ammonia molecules ≈ 4.5865 mol x (\(6.022 * 10^{23\) molecules/mol)

                           ≈ \(2.76 * 10^{24\) molecules (rounded to two significant figures)

Therefore, the provided sample contains roughly \(2.76 * 10^{24\) ammonia molecules.

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The number of ammonia molecules in the sample is approximately 2.764 x \(10^{24}\) molecules.

To calculate the number of ammonia molecules in a given sample, we need to use Avogadro's number and the molar mass of ammonia.

The molar mass of ammonia (NH3) is calculated as follows:

Molar mass of N = 14.01 g/mol

Molar mass of H = 1.01 g/mol

Total molar mass of NH3 = 14.01 g/mol + (3 * 1.01 g/mol) = 17.03 g/mol

Now, we can calculate the number of moles of ammonia in the sample:

Number of moles = Mass of sample / Molar mass of NH3

Number of moles = 78.25 g / 17.03 g/mol = 4.594 moles

Next, we use Avogadro's number, which states that there are 6.022 x \(10^{23}\) molecules in one mole of a substance.

Number of molecules = Number of moles * Avogadro's number

Number of molecules = 4.594 moles * 6.022 x \(10^{23}\) molecules/mol = 2.764 x \(10^{24}\) molecules

Therefore, there are approximately 2.764 x \(10^{24}\) ammonia molecules in the given sample of 78.25 g.

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Answer:

B

Explanation:

Answer:

Ca2+

Explanation:

don't really know what to explain here, thats just the ion name.

The arrow in a chemical equation represents the direction of the reaction. It indicates the conversion of reactants into products. The arrow points from the reactant side to the product side, symbolizing the flow of the reaction.

The purpose of the arrow is to visually represent the chemical transformation occurring in the reaction. It shows the relationship between the reactants and products and the direction in which the reaction proceeds. The arrow implies that the reactant molecules are being rearranged and transformed into new substances with different properties.

Chemical equations are used to describe the stoichiometry and balance of reactions. The arrow helps convey this information by illustrating the overall process taking place. It serves as a crucial element in understanding the reaction's composition, reaction conditions, and the substances involved.

Furthermore, the arrow also implies that the reaction can occur in both directions. In reversible reactions, the arrow can be represented as a double-headed arrow, indicating that the reaction can proceed in either direction depending on the conditions.

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Answer:

12.5

Explanation:

After three half-lives, one eighth (1/2 x 1/2 x 1/2) of the uranium-238 will remain. One-eighth equals 12.5%