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What Is the noble gas configuration for strontium (atomic number = 38)? Given: He (atomic number = 2) Ne (atomic number = 10) Ar (atomic number = 18) Kr (atomic number = 36) B В A [Kr 45Kr 5​

Answer:

Answer is A.

Explanation:

The indivisibility of an atom was proved wrong: an atom can be further subdivided into protons, neutrons and electrons. According to Dalton, the atoms of same element are similar in all respects. However, atoms of some elements vary in their masses and densities. These atoms of different masses are called isotopes. :)

Answer:

Explanation:

418.4kj is the correct answer

Answer:

Nitrobenzene < Bromobenzene < Benzene < Phenol

Explanation:

Aromatic compounds undergo electrophilic aromatic substitution reaction in the presence of relevant electrophiles. Certain substituents tend to increase or decrease the tendency of an aromatic compound towards electrophilic aromatic substitution reaction.

Substituents that increase the electron density around the ring such as in phenol tends to make the ring more reactive towards electrophilic substitution. Halogens such as bromine has a -I inductive effect as well as a +M mesomeric effect.

However the -I(electron withdrawing effect) of the halogens supersedes the +M electron donation due to mesomeric effect.

Putting all these together, the order of increasing reactivity of the compounds towards electrophilic aromatic substitution is;

Nitrobenzene < Bromobenzene < Benzene < Phenol

385 x 42.13 x 0.079 = 1281.38395

Answer:

Choice A. [tex]\rm H_{2}[/tex] would be produced at the negative electrode.

Explanation:

Ionic equation for this reaction:

[tex]2\, {\rm K^{+}} + 2\, {\rm Cl^{-}} + {2\, \rm H_{2} O} \to {\rm H_{2}} + {\rm Cl_{2}} + 2\, {\rm K^{+}} + {\rm 2\, OH^{-}}[/tex].

Net ionic equation:

[tex]2\, {\rm Cl^{-}} + 2\, \rm H_{2} O} \to {\rm H_{2}} + {\rm Cl_{2}} + 2\, {\rm OH^{-}}[/tex].

Half-equations:

[tex]2\, {\rm Cl^{-}} \to {\rm Cl_{2}} + 2\, {e^{-}}[/tex].

(Electrons travel from the solution to an electrode.)

[tex]2\, {\rm \overset{+1}{H}_{2} O} + 2\, {e^{-}} \to \overset{0}{\rm H}_{2} + 2\, {\rm O\overset{+1}{H}\!^{-}}[/tex].

(An electrode supply electrons to the solution to reduce some of the [tex]\rm H[/tex] atoms from [tex]\rm H_{2}O[/tex].)

In a DC circuit, electrons always enter the circuit from the negative terminal of the power supply and return to the power supply at the positive terminal.

The negative electrode is connected to the negative terminal of the power supply. Electrons from the power supply would flow into the solution through this electrode.  

This continuous supply of electrons at the negative electrode would drive a reduction half-reaction. In this question, that corresponds to the reduction of water: [tex]2\, {\rm \overset{+1}{H}_{2} O} + 2\, {\rm e^{-}} \to \overset{0}{\rm H}_{2} + 2\, {\rm O\overset{+1}{H}\!^{-}}[/tex]. Hence, [tex]\rm H_{2}[/tex] would be produced at the negative electrode.

Answer:

+0.151, spontaneous

Explanation:

Given that;

Co^2+(aq) + 2e ---->Co(s) -0.28 V

Pb^2+(aq) + 2e ---->Pb(s). -0.13 V

Hence Co is the anode and Pb is the cathode

E°cell = E°cathode - E°anode

So;

E°cell = -0.13 V - (-0.28 V)

E°cell = 0.15 V

The cell reaction is spontaneous since E°cell is positive.

Kc = [CH4]×[H2O] / [CO]×[H2]^3

Kc = 1.078×0.878 / (0.989×0.933^3)

Kc = 0.977

The numerical value of the equilibrium constant, Kc, for the given reaction is found to be 0.977.

The Equilibrium constant may be defined as the numerical value that significantly indicated the correlation between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a definite temperature.

According to the question, the reaction is as follows:

[tex]CO +3H_2[/tex] ↔ [tex]CH_4+ H_2O[/tex].

The equilibrium concentrations are 0.989 M, 0.993 M, 1.078 M and 0.878 M, respectively.

Now, the equilibrium constant is calculated by the following formula:

             =  1.078×0.878 / (0.989×0.93[tex]3^3[/tex]).

             = 0.9464/(0.989 × 0.8121)

             =  0.977.

Therefore, the numerical value of the equilibrium constant, Kc, for the given reaction is found to be 0.977.

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Answer:

cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3

Answer:

Specific Heat Capacity

Answer:

Explanation:

Na2CO3+Ca(NO3)2=CaCO3+2NaNO3

nNa2CO3=0.02

nCa(NO3)2=0.02

mCaCO3=0.02*100=2 gram

nNaNo3=0.04

Cm=2/15

From the calculation, the mass of the product is 2 g.

A chemical reaction occurs when two more substances are mixed together. In this case, the reaction is shown by; Ca(NO3)2 + Na2CO3 ----> CaCO3(s) + 2NaNO3.

Number of moles of Na2CO3 = 100/1000 L *  0.2 mol/L = 0.02 moles

Number of moles of  Ca(NO3)2 = 200/1000 L *  0.1 mol/L = 0.02 moles

Since the reaction is equimolar, amount of the product = 0.02 moles * 100 g/mol = 2 g

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Answer:

Sorry but i don't undertsnad the question.

Explanation:

Answer:

A physical change is a change to the physical—as opposed to chemical—properties of a substance. They are usually reversible. The physical properties of a substance include such characteristics as shape, color, texture, flexibility, density, and mass.

A chemical change happens when one chemical substance is transformed into one or more different substances, such as when iron becomes rust.

Do u want examples ?

Solution :

We know that :

[tex]$\Delta T_f = k_f.m$[/tex]  and   [tex]$m=\frac{w_2}{m_2 \times w_1}$[/tex]

Then, [tex]$\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$[/tex]   ..................(1)

Where,

[tex]w_1[/tex] = amount of solvent (in kg)

[tex]w_2[/tex] = amount of solute (in kg)

[tex]m_2[/tex] = molar mass of solute (g/mole)

[tex]m[/tex] = molality of solution (mole/kg)

Given :

[tex]\Delta T_f[/tex] = [tex]3.14\ ^\circ C[/tex],   [tex]k_f= 5.12\ ^\circ C/m[/tex]

                              [tex]=5.12 \ ^\circ C/mole/kg[/tex]

                              [tex]=5.12 \ ^\circ C \ kg/mole[/tex]

[tex]w_1[/tex] = 0.250 kg,  [tex]w_2[/tex] = 24.3 g

Then putting this values in the equation is (1),

[tex]$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$[/tex]

[tex]$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$[/tex]

[tex]m_2= 158.49[/tex]  g/mole

So, the molar mass of the unknown compound is 158.49 g/mole.

Answer:

Covalent bonding and non-covalent bonding

Answer:

Hence the solids that should test to investigate the effects of cations on pH is

[tex]AlBr_{3}[/tex] (Cation is Al 3+)  

[tex]MgCl_{2}[/tex]  ( Cation is Mg 2+)  

[tex]CH_{3} NH_{3} Br[/tex] ( Cation is NH2+).

Explanation:

The solids in the following should you test to investigate the effects of cations on pH.  

[tex]AlBr_{3}[/tex] contains (Cation is Al 3+)  

[tex]MgCl_{2}[/tex] contains ( Cation is Mg 2+)  

[tex]CH_{3} NH_{3} Br[/tex] contains( Cation is NH2+ )

The atoms or the molecules containing the positive charge that gets attracted to the cathode are called cations. The compounds a. [tex]\rm AlBr_{3}[/tex], c. [tex]\rm MgCl_{2}[/tex] and e. [tex]\rm CH_{3}NH_{3}Br[/tex] should be investigated.

Cations are the positive charge containing molecules and atoms that have more protons in their nucleus than the number of electrons in their shells. They are formed when they lose one or more electrons to another atom.

The addition or release of the electrons of the cations and anions affects the pH system as absorption of the cation decreases the pH and absorption of the anions increases the pH.

Hence, [tex]\rm Al^{3+}[/tex], [tex]\rm Mg^{2+}[/tex] and [tex]\rm NH^{2+}[/tex] are the cation that should be investigated. The addition of the cations will reduce the pH of the reaction.

Therefore, absorption of the cation reduces the pH.

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Answer:

Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?

A. HBr, Na2S, Mg(OH)2, Na2CO3.

B. H2SO4, NaOH, NaCl, HF.

C. HNO3, H2SO4, KOH, K2SiO3.

D. Ca(OH)2, KOH, CH3COOH, NaCl.

Answer:

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Explanation:

We will balance the redox reaction through the ion-electron method.

Step 1: Identify both half-reactions

Reduction: Br₂ ⇒ Br⁻

Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻

Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate

Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺

Step 3: Perform the charge balance, adding electrons where appropriate

2 e⁻ + Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻

Step 4: Make the number of electrons gained and lost equal

5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)

1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)

Step 5: Add both half-reactions

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Answer:

Anaxagoras was perhaps the first literate person to attempt to explain physical phenomena rationally, basing his ideas upon careful observations and simple experiments. This is fundamental to modern science and is the sine qua non of environmental study.

Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.

1.42liters, which is equivalent to 3pints, of blood is required for the surgery

1 US pint = 0.473 liters

3 × 0.473

= 1.419 liters of blood for the surgery

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Answer:

Solution given:

1 mole of KCl[tex]\rightarrow [/tex]22.4l

1 mole of KCl[tex]\rightarrow [/tex]74.55g

we have

0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g

74.55g of KCl[tex]\rightarrow [/tex]22.4l

10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.

[tex]\:[/tex]

1 mole of KCl → 22.4l

1 mole of KCl → 74.55g

we have

0.14 mole of KCl → 74.55*0.14=10.347g

74.55g of KCl  → 22.4l

10.347 g of KCl → 22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.

[tex]\\ \large\sf\longmapsto KNO_3[/tex]

[tex]\\ \large\sf\longmapsto 39u+14u+3(16u)[/tex]

[tex]\\ \large\sf\longmapsto 53u+48u[/tex]

[tex]\\ \large\sf\longmapsto 101u[/tex]

[tex]\\ \large\sf\longmapsto 101g/mol[/tex]

Now

[tex]\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}}[/tex]

[tex]\\ \large\sf\longmapsto No\:of\:moles=\dfrac{0.565}{101}[/tex]

[tex]\\ \large\sf\longmapsto No\:of\:moles=0.005mol[/tex]

We know

[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Vol\:of\:Solution\:in\:L}}[/tex]

[tex]\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{\dfrac{250}{1000}L}[/tex]

[tex]\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{0.250}[/tex]

[tex]\\ \large\sf\longmapsto Molarity=0.02M[/tex]

[tex] \: \: \: \: \: \: \: \: \: [/tex]

Answer:

volume versus temperature, because they are also directly proportional.

Explanation:

Just took the test!

Answer:

Explanation:

A balanced chemical equation refers to the reaction taking place whereby the number of atoms associated in the reactants side is equivalent to the number of atoms on the products side.

From the given information, the balanced equations are as follows:

[tex]\mathbf{(a) \ \ \ Ti(s) + O_{2(g)} \to TiO_{2(s)}}[/tex]

[tex]\mathbf{(b) \ \ \ 2Ag_{2}O \to 4Ag_{(s)} + O_{2(g)}}[/tex]

[tex]\mathbf{(c) \ \ \ 2C_3H_7OH + 9O_2 \to 6CO_2+8H_2O}[/tex]

[tex]\mathbf{(d) \ \ \ 2C_5 H_{12}O \to 10 CO_2 + 12 H_2O}[/tex]

Explanation:

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The equilibrium constant for the reaction is K = 0.137

We obtain the equilibrium constant considering the following equilibria and their constants:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

COO(s) + CO(g) → Co(s) + CO₂(g)   K₂ = 490

We write the first reaction in the forward direction because we need H₂(g) in the reactants side:

(1)     COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):

(2)   Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

From the addition of (1) and (2), we obtain:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

+

Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

-------------------------------------------------

H₂(g) +  CO₂(g) → CO(g) + H₂O(g)

Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.

Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:

K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137

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Answer:

Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.

Number of half lives in 9612 years = 9612/1602 = 6 half lives

New mass = Original mass x (1/2)n where n is the number of half lives.

Therefore, New mass= 500 x (1/2)∧6

                                 = 500 x 0.015625

                                 = 7.8125 g

Hence the mass of radium after 9612 years will be 7.8125 grams.

Explanation:

Answer:

[tex]\boxed {\boxed {\sf 6.25 \ grams}}[/tex]

Explanation:

We are asked to find the mass of a sample of radium-226 after half-life decay. We will use the following formula:

[tex]A= A_o *\frac{1}{2}^{\frac{t}{h}}[/tex]

In this formula, [tex]A_o[/tex] is the initial amount, t is the time, and h is the half-life.

For this problem, the initial amount is 200 grams of radium-226, the time is 8,000 years, and the half-life is 1,600 years.

[tex]\bullet \ A_o= 200 \ g \\\\bullet \ t= 8,000 \ \\\bullet \ h= 1,600[/tex]

Substitute the values into the formula.

[tex]A= 200 \ g * \frac{1}{2} ^{\frac{8.000}{1,600}[/tex]

Solve the fraction in the exponent.

[tex]A= 200 \ g * \frac{1}{2}^{5}[/tex]

Solve the exponent.

[tex]A= 200 \ g *0.03125[/tex]

[tex]A= 6.25 \ g[/tex]

In addition, we can solve this another way. First, we find the number of half-lives by dividing the total time by the half-life.

Every half-life, 1/2 of the mass decays. Divide the initial mass in half, then that result in half, and so on 5 times.

After 8,000 years, 6.25 grams of radium-226 remains.

Answer:

Biphenyl

Explanation:

The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.

The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.

Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.

Answer:

Radical chain initiator

Explanation:

The peroxide here serves as a radical chain initiator. In the field of chemistry the radical initiatives are those substances that are used in industrial processes like polymer synthesis. These initiatives have weak bonds generally and they're mostly used to create free radicals. These radicals are atoms that have odd numbers of electrons. Peroxide is an example of such.