why was there no reaction seen between barium nitrate and sodium chloride?

1) The boron atom has only one unpaired electron, making it paramagnetic.

2) Boron atom has one unpaired electron, making it paramagnetic.Iron atom has four unpaired electrons, making it paramagnetic.

1. Boron atom and unpaired electrons Boron is a chemical element with the symbol B and atomic number 5. It is a trivalent metalloid and has three valence electrons. The electron configuration of boron is 1s² 2s² 2p¹. Therefore, the boron atom has only one unpaired electron, making it paramagnetic.2. Iron atom and unpaired electronsIron is a chemical element with the symbol Fe and atomic number 26. It is a metal and has two valence electrons. The electron configuration of iron is [Ar] 3d⁶ 4s². Therefore, the iron atom has four unpaired electrons, making it paramagnetic. Answer:Boron atom has one unpaired electron, making it paramagnetic.Iron atom has four unpaired electrons, making it paramagnetic.

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The concentration of calcium ion in the pill is 18,320 ppm.

To calculate the concentration of calcium ion in parts per million (ppm), we need to determine the mass of calcium ion in the pill and divide it by the mass of the pill, then multiply by 1,000,000.

Mass of the pill = 2.43 g

Mass of Ca^2+ = 44.6 mg = 0.0446 g

Now, we can calculate the concentration in ppm:

Concentration of Ca^2+ (ppm) = (Mass of Ca^2+ / Mass of the pill) * 1,000,000

Concentration of Ca^2+ (ppm) = (0.0446 g / 2.43 g) * 1,000,000

Concentration of Ca^2+ (ppm) ≈ 18,320 ppm

The concentration of calcium ion in the pill is approximately 18,320 ppm.

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Answer:

July 23

Explanation:

The correct answer is c. C₂H₂, which has the shortest carbon-carbon bond length.

The compound with the shortest carbon-carbon bond length is c. C₂H₂, which refers to ethyne or acetylene. Ethyne consists of a triple bond between the two carbon atoms, resulting in a shorter bond length compared to the other compounds listed.

In option a, CH₃CH₃ (ethane), the carbon-carbon bond is a single bond, which is longer than a triple bond.

In option b, CH₂CH₂ (ethylene), the carbon-carbon bond is a double bond, which is longer than a triple bond but shorter than a single bond.

Therefore, the correct answer is c. C₂H₂, which has the shortest carbon-carbon bond length due to the presence of a triple bond between the carbon atoms.

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The delta H for the reaction can be calculated using average bond energies. Based on the provided table, the bond energies are as follows: C-C (839 kJ/mol), H-I (348 kJ/mol), C-I (299 kJ/mol), and C-H (240 kJ/mol). To determine the delta H, we need to subtract the sum of the bond energies broken from the sum of the bond energies formed.

In this case, the bonds broken are C-C and H-I, with energies of 839 kJ/mol and 348 kJ/mol respectively. The bonds formed are C-I and C-H, with energies of 299 kJ/mol and 240 kJ/mol respectively. Calculating the delta H: Delta H = (Energy of bonds broken) - (Energy of bonds formed) = (839 kJ/mol + 348 kJ/mol) - (299 kJ/mol + 240 kJ/mol) = 1187 kJ/mol - 539 kJ/mol = 648 kJ/mol. Therefore, the delta H for the reaction is +648 kJ/mol. Since none of the given options match this value exactly, it seems there may be an error in the options provided.

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1a. The balanced chemical equation for the reaction of aqueous lead (II) nitrate [tex]\ce{Pb(NO3)2}[/tex] and potassium chromate [tex]\ce{K2CrO4}[/tex] can be written as follows:

[tex]\ce{Pb(NO3)2(aq) + K2CrO4(aq) - > PbCrO4(s) + 2KNO3(aq)}[/tex]

1b. Approximately 1.9392 grams of lead chromate are formed from the reaction.

To determine the number of grams of lead chromate [tex]\ce{PbCrO4}[/tex] formed from the reaction of 15.0 mL of 0.40 M potassium chromate ([tex]\ce{K2CrO4}[/tex]) with 15 mL of lead nitrate [tex]\ce{Pb(NO3)2}[/tex], we need to first find the limiting reagent and then calculate the amount of lead chromate produced.

First, let's calculate the number of moles of potassium chromate:

Moles of [tex]\ce{K2CrO4}[/tex]= (0.40 mol/L) × (0.015 L) = 0.006 mol

Next, let's calculate the number of moles of lead nitrate:

Moles of [tex]\ce{Pb(NO3)2}[/tex] = (0.40 mol/L) × (0.015 L) = 0.006 mol

From the balanced equation, the stoichiometric ratio between [tex]\ce{K2CrO4}[/tex]and [tex]\ce{PbCrO4}[/tex] is 1:1. It means that 1 mole of K2CrO4 reacts with 1 mole of [tex]\ce{PbCrO4}[/tex].

Therefore, the number of moles of[tex]\ce{PbCrO4}[/tex] formed is 0.006 moles.

To find the mass of [tex]\ce{PbCrO4}[/tex], we need to multiply the number of moles by its molar mass. The molar mass of [tex]\ce{PbCrO4}[/tex]is:

Molar mass of [tex]\ce{PbCrO4}[/tex]= (207.2 g/mol) + (52.0 g/mol) + (4 × 16.0 g/mol) = 323.2 g/mol

Mass of [tex]\ce{PbCrO4}[/tex]= 0.006 mol × 323.2 g/mol = 1.9392 g

Therefore, approximately 1.9392 grams of lead chromate are formed from the reaction.

1c. Since the stoichiometric ratio between [tex]\ce{K2CrO4}[/tex] and [tex]\ce{PbCrO4}[/tex] is 1:1, and the number of moles of [tex]\ce{K2CrO4}[/tex] and [tex]\ce{PbCrO4}[/tex] are equal (0.006 moles), neither of them is the limiting reagent. In this case, both reactants are present in excess, and there is no limiting reagent.

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Answer:

Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent like adding more water to the solution

Answer:

ions

Explanation:

Answer:

I think B

Explanation:

Red litmus paper is used more to find bases, blue litmus paper changes better when coming to find acids. So I believe it is B

The best approach best investigate whether substances are basic is to test each substance with red litmus paper.

Basic compounds are those which have a pH value in between the range of 7.1 to 14.

Red litmus paper changes its color from red to blue when comes in the contact of basic compounds. So to check the basicity of household substance we should use red litmus paper.

Hence test each substance with red litmus paper.

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If the temperature were lowered to 100∘c, the rate of the forward reaction would decrease.

Ammonia (NH3) is commonly used in various industrial processes such as the synthesis of pharmaceuticals like antimalarials and vitamins like B vitamins.

The equation for the production of ammonia is N2(g) + 3H2(g) ⇌ 2NH3(g). The Haber process is carried out at a temperature of about 500°C.

If the temperature is decreased to 100°C, the rate of the forward reaction will decrease. In other words, the equilibrium position will shift in the direction of the reverse reaction. The decrease in temperature will lower the kinetic energy of the reactant molecules, thus reducing their rate of collision and therefore decreasing the rate of the forward reaction.

:If the temperature were lowered to 100∘c, the rate of the forward reaction would decrease.

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The pH of the buffer after the addition of 0.068 mol NaOH is calculated to be approximately 11.02.

To determine the pH of the buffer after the addition of NaOH, we need to consider the reaction that occurs between NaOH and the components of the buffer, which are HBr and ethylamine C₂H₅NH₂.

The reaction between NaOH and HBr is a neutralization reaction:

HBr + NaOH → NaBr + H₂O

This reaction consumes HBr and produces water. The reaction between NaOH and ethylamine is an acid-base reaction:

NaOH + C₂H₅NH₂ → NaC₂H₅NH₂ + H₂O

This reaction consumes NaOH and produces ethylamine salt (sodium ethylamine) and water.

Given that 0.068 mol of NaOH is added, we need to determine which component of the buffer is limiting and calculate the remaining amounts of each component.

moles of HBr = (0.0805 L) x (1.05 mol/L) = 0.084525 mol

moles of ethylamine = (0.2049 L) x (0.953 mol/L) = 0.1955097 mol

Given the moles NaOH = 0.068 mol

Since the moles of HBr (0.084525 mol) is greater than the moles of NaOH (0.068 mol), HBr is the limiting component.

So, the remaining moles HBr

= moles HBr - moles NaOH

= 0.084525 mol - 0.068 mol

= 0.016525 mol

Volume of the buffer = 0.0805 L + 0.2049 L = 0.2854 L

The final concentration of HBr

= remaining moles HBr / volume of the buffer

= 0.016525 mol / 0.2854 L ≈ 0.0579 M

So, pOH

=[tex]-log_{10}(Kb) + log_{10}(concentration of the ethylamine)[/tex]

= [tex]-log10(4.5 \times 10^{-4}) + log_{10}(0.953 M)[/tex]

≈[tex]-log10(4.5 \times 10^{-4}) + 0.9793[/tex]

pH = 14 - pOH

  ≈ [tex]14 - (-log_{10}(4.5 \times 10^{-4}) + 0.9793) \approx 11.2[/tex]

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The standard cell potential (E°cell) of the reaction is found to be + 0.38 V, hence, option A is correct.

We will be usig the equation,

E°cell = (ΔG° / -nF) standard Gibbs free energy change is ΔG°, number of electrons transferred in the balanced equation is n, and Faraday constant (96,500 C/mol) is F. In the given reaction, 3 electrons are transferred, so n = 3.

Given ΔG° = -110 kJ/mol and F = 96,500 C/mol, we can substitute these values into the equation to calculate E°cell,

E°cell = (-110,000 J/mol / (-3 * 96,500 C/mol))

E°cell = 0.38 V

Therefore, the answer is, a. + 0.38 V

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In the direct reaction HCO3- + H2O ⇌ CO32- + H3O+, HCO3- acts as a base.

H2PO4- and HPO42-

H3O+ and OH-

H2CO3 and CO32 are

What is conjugate acid?

An acid and a base which differ only by the presence or absence of a proton are called a conjugate acid-base pair.

Part A:

In the direct reaction HCO3- + H2O ⇌ CO32- + H3O+, HCO3- acts as a base. This is because it accepts a proton (H+) from water (H2O) to form H3O+ (a hydronium ion). In this reaction, HCO3- acts as a Bronsted-Lowry base.

Part B:

Conjugate acid/base pairs among the options are:

H2PO4- and HPO42- (acid/base conjugate pair)

H3O+ and OH- (acid/base conjugate pair)

HCl and Cl- (not a conjugate acid/base pair; both are ions but not related by proton transfer)

H2CO3 and CO32- (acid/base conjugate pair)

So the correct answers are:

H2PO4- and HPO42-

H3O+ and OH-

H2CO3 and CO32-

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Answer:

The reaction is exothermic as the solution becomes warm.

Explanation:

Hello there!

In this case according to the described reaction between sodium metal and water, we can write up the chemical equation whereas the products turn out to be sodium hydroxide and hydrogen gas:

[tex]2Na+2H_2O\rightarrow 2NaOH+H_2[/tex]

Moreover, these reactions are known to be highly exothermic, for that reason the solution becomes warm as the reaction releases heat as it goes to completion.

Best regards!

Answer:

The answer is C

It's a model of Decomposition reaction

Answer:

A is endothermic, B is exothermic.

Explanation:

I actually just looked at the bottom of the graph. Endothermic reactions absorb energy, but exothermic reactions release it. Hope this helped!

Answer:

habitat destruction, overexploitation, climate change, nitrogen pollution, and invasive species.

Explanation:

Main Modern Causes of Extinction:

habitat destruction - the process by which a natural habitat becomes incapable of supporting its native species.

overexploitation - harvesting a renewable resource to the point of diminishing returns.

climate change - includes both global warmings driven by human emissions of greenhouse gases and the resulting large-scale shifts in weather patterns.

nitrogen pollution - a form of water pollution, refers to contamination by excessive inputs of nutrients.

invasive species - an introduced organism that negatively alters its new environment.

Answer: Communicable diseases, abuse gunshot wounds and forensic medicine

Explanation:

The element with the greatest first ionization energy among the given options is Se (selenium).

Option (A) is correct.

The first ionization energy refers to the energy required to remove one electron from an atom in its neutral state, forming a positively charged ion. The greater the ionization energy, the more difficult it is to remove an electron.

Considering the elements provided, analyze their positions in the periodic table to make an educated guess:

A. Se (selenium) - Selenium is found in Group 16 (Group 6A) of the periodic table.

B. S (sulfur) - Sulfur is also found in Group 16 (Group 6A) of the periodic table.

C. K (potassium) - Potassium is found in Group 1 (Group 1A) of the periodic table.

D. Cl (chlorine) - Chlorine is found in Group 17 (Group 7A) of the periodic table.

E. Ca (calcium) - Calcium is found in Group 2 (Group 2A) of the periodic table.

Based on the periodic trends, the elements in the upper right portion of the periodic table tend to have the greatest first ionization energies. This is because these elements have a higher effective nuclear charge and a smaller atomic radius.

Comparing the given options, we can see that:

A. Se and B. S are both in Group 16 (Group 6A). Since they are closer to the upper right portion of the periodic table, would expect them to have higher first ionization energies compared to the other options.

C. K is in Group 1 (Group 1A), which is in the far left portion of the periodic table. Elements in this group tend to have lower first ionization energies compared to those in the upper right portion.

D. Cl is in Group 17 (Group 7A), which is closer to the upper right portion of the periodic table compared to Group 1. Therefore, chlorine would have a higher first ionization energy than potassium but likely lower than selenium and sulfur.

E. Ca is in Group 2 (Group 2A), which is to the left of Group 1. Elements in Group 2 have higher first ionization energies compared to those in Group 1 but generally lower than elements in the upper right portion.

Considering these trends, the element with the greatest first ionization energy among the given options is:

A. Se (selenium)

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Answer:

To make glue. aaaaaaaaa

Answer:

Cleaning

Explanation:

Hydrochloric acid is an ingredient in household cleaners because it helps clean tough stains.

The balanced equation is 6 Pb(s) + 6 Cr2O72-(aq) + 42 H+(aq) → 12 Cr3+(aq) + 42 H2O(l) + 6 Pb2+(aq). 6 moles of Pb(s) are oxidized for every 6 moles of Cr2O72- consumed. If 5 moles of Cr2O72- are used, it will oxidize 5 moles of Pb(s).

First, we need to balance the two half-reactions. Let's start with the reduction half-reaction of Cr2O72-: Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l).  To balance the oxygen atoms, we add 7 H2O to the left side: Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l). Next, we balance the hydrogen atoms by adding 14 H+ to the right side:

Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l) + 14 H+(aq)

Now, let's balance the oxidation half-reaction of Pb: Pb(s) → Pb2+(aq) + 2 e-. The equation is already balanced. To combine the two half-reactions, we multiply the reduction half-reaction by 6 and the oxidation half-reaction by 6 to ensure that the electrons cancel out: 6 Cr2O72-(aq) + 42 H+(aq) + 36 e- → 12 Cr3+(aq) + 42 H2O(l) + 84 H+(aq). 6 Pb(s) → 6 Pb2+(aq) + 12 e-. Now, we can write the balanced equation: 6 Pb(s) + 6 Cr2O72-(aq) + 42 H+(aq) → 12 Cr3+(aq) + 42 H2O(l) + 6 Pb2+(aq).

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Answer:

(A) 0.80 M

(B) 0.64 g

(C) 0.050 M

(D) 1.6 M

Explanation:

(A)

Convert 4.0 g to moles with molar mass. Convert 125 mL to L.

4.0 g x (1 mol/39.998 g) = 0.100 mol

125 mL x (1 L/1000 mL) = 0.125 L

Molarity = mol/L

Molarity = 0.100 mol/0.125 L = 0.80 M

(B)

Find moles of NaOH with the molarity you just found. And convert 20 mL to L.

Molarity = mol/L

0.80 M = mol/0.020 L

mol = 0.016

Convert moles to grams with molar mass.

0.016 mol x (39.998 g/1 mol) = 0.64 g

(C)

You know the molarity of a NaOH solution (Part A). 30 mL initial volume and 480 mL final volume are given. Dilution question so use the dilution formula.

M1V1 = M2V2

(0.80 M)(30 mL) = (M2)(480 mL)

M2 = 0.050 M

(D)

Dilution question again, use the dilution formula.

M1V1 = M2V2

(0.80 M)(200 mL) = (M2)(100 mL)

M2 = 1.6 M

Answer:To be honest i dont know

Explanation:

e

Answer:

Explanation:

Alpha radiation ionizes the air. When smoke interaccts with the ionized particles it causes the alarm to sound.

Answer:

The period number tells which is the highest energy level occupied by the electrons.

Explanation:

Answer:

Percent volume-volume (%(v/v)) = 100 x (volume of solute / volume of solution)

Percent volume-volume (%(v/v)) = 100 x (volume of solute / volume of solution)ex. 20 ml of methanol dissolved in enough water to make 200 ml of solution would result in a 10 % methanol solution

Percent volume-volume (%(v/v)) = 100 x (volume of solute / volume of solution)ex. 20 ml of methanol dissolved in enough water to make 200 ml of solution would result in a 10 % methanol solution-the units may be any units of volume you chose - as long as they are consistent

Percent volume-volume (%(v/v)) = 100 x (volume of solute / volume of solution)ex. 20 ml of methanol dissolved in enough water to make 200 ml of solution would result in a 10 % methanol solution-the units may be any units of volume you chose - as long as they are consistent-this concentration unit is most often used when mixing two liquids

Answer:

[tex]\boxed {\boxed {\sf 9.6 \ mol \ KCl}}[/tex]

Explanation:

We must use stoichiometry to solve this, which is the calculation of reactants and products in a reaction using ratios.

Let's analyze the reaction given.

[tex]MgCl_2 _{(aq)} + K_2SO_4 _{(aq)} \rightarrow 2KCl _{(aq)} + MgSO_4 _{(s)}[/tex]

Now, look at the coefficients, or numbers in front of the molecule formulas. If there isn't a coefficient, then a 1 is implied.

We want to find how many moles of potassium chloride (KCl) are produced from 4.8 moles of magnesium chloride (MgCl₂). Check the coefficients for these molecules.

The coefficient represents the number of moles. Therefore, 1 mole of magnesium chloride produces 2 moles of potassium chloride. We can set up a ratio using this information.

[tex]\frac { 1 \ mol \ MgCl_2} {2 \ mol \ KCl}[/tex]

Multiply by the given number of moles of magnesium chloride: 4.8

[tex]4.8 \ mol \ MgCl_2 *\frac { 1 \ mol \ MgCl_2} {2 \ mol \ KCl}[/tex]

Flip the ratio so the moles of magnesium chloride cancel out.

[tex]4.8 \ mol \ MgCl_2 *\frac {2 \ mol \ KCl} { 1 \ mol \ MgCl_2}[/tex]

[tex]4.8 *\frac {2 \ mol \ KCl} { 1 \ } }[/tex]

[tex]4.8 * {2 \ mol \ KCl}[/tex]

[tex]9.6 \ mol \ KCl[/tex]

9.6 moles of potassium chloride are produced from 4.8 moles of magnesium chloride.

Answer:

N with a charge of (-3) is nitride

Explanation: