The compound chromium(II) chloride is a strong electrolyte. Write the transformation that occurs when solid chromium(II) chloride dissolves in water. Be sure to specify states such as (aq) or (s).

The correct answer is C. square planar

According to the Valence Shell Electron Pair Repulsion Theory(VSEPR), The shape of a molecule depends on the number of electron pairs in the molecule.

VSEPR theory was first coined by Gillespie  and Nyhlom in 1957 as an improvement over the Sidgwick - Powell theory.

According to this theory, the shape of a molecule is determined by the number of electron pairs that surround the valence shell of the central atom in the molecule.  The electron pairs are positioned as far apart in space as possible to minimize repulsion of electron pairs.

However, the presence of lone pairs distorts the shape anticipated for the molecule on the basis of VSEPR.

For a molecule having six electron pairs, an octahedral geometry is expected(electron domain geometry). However, the presence of two lone pairs which are positioned at opposite side of the four single bonds leads to an observed square planar molecular geometry.

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Answer:

square planar

Explanation:

Answer:

10 ms-1

Explanation:

Kinetic energy = 1/2 × m × v^2

1 = 1/2× 20 ×10^ -3 × v^2

v ^ 2 = 100

v = 10 ms-1

note : convert grams in to kg before substitution as above

Given:

Kinetic energy,

Mass,

We know the formula,

→ [tex]K.E = \frac{1}{2} mv^2[/tex]

By putting the values, we get

       [tex]1 = \frac{1}{2}\times 20\times 10^{-3}\times (v)^2[/tex]

     [tex]v^2 = 100[/tex]

       [tex]v = \sqrt{100}[/tex]

       [tex]v = 10 \ m/s[/tex]

Thus the above response is correct.

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Answer:A) V = 16.892 ml

Explanation:

M1 * V1 = M2 * V2

14.8 M * V1 =0.250 M * 1000 ml

V1 = 16.892 ml

a. The volume of 16.89 milliliters of the stock solution of 14.8 M  should be diluted to make 1000.0 mL of 0.250 M.

b. The concentration of the final solution is 0.296 M.

The concentration or the volume of the concentrated or dilute solution can be calculated by using the equation:

M₁V₁ = M₂V₂

where M₁ and V₁ are the concentration and volume of the concentrated solution respectively and M₂ and V₂ are the concentration and volume of the dilute solution.

A stock solution is a solution that has a high concentration and that will be diluted to a low concentration by the addition of water in it.

Given, a stock solution of concentration, M₁ = 14.8 M

The concentration of the diluted solution, M₂ = 0.250 M

The volume of diluted solution, V₂  = 1000ml

Substitute the value of the molarity and volume in equation (1):

(14.8)× (V₁) = (1000) × (0.250)

V₁ = 16.89 ml

Similarly, for part (b): M₁ = 14.8 M, V₁ = 10 ml and V₂  = 0.5L = 500 ml

(14.8)× (10) = (500) × (M₂)

M₂ = 0.296 M

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Answer:

17

Explanation:

Step 1: Calculate the needed concentrations

[A]i = 1.00 mol/5.00 L = 0.200 M

[B]i = 1.80 mol/5.00 L = 0.360 M

[B]e = 1.00 mol/5.00 L = 0.200 M

Step 2: Make an ICE chart

        A(aq) + 2 B(aq) ⇄ C(aq)

I       0.200    0.360        0

C        -x           -2x         +x

E     0.200-x  0.360-2x   x

Then,

[B]e = 0.360-2x = 0.200

x = 0.0800

The concentrations at equilibrium are:

[A]e = 0.200-0.0800 = 0.120 M

[B]e = 0.200 M

[C]e = 0.0800 M

Step 3: Calculate the concentration equilibrium constant (K)

K = [C] / [A] × [B]²

K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17

Answer:

B. 0.0000005461m

I used the method of moving the decimal.

Answer:

25.88 g/mol

Explanation:

Graham's law is a famous law which states that the diffusion rate or the effusion rate of any gas varies inversely to the square root of the molecular weight the gas.

So from Graham's law, we have,

[tex]$\frac{\text{time}}{M^{1/2}}=\text{constant}$[/tex]

Using the sample of Kr gas having M = 83.8

[tex]$\frac{8.15}{(83.8)^{0.5}}= \frac{4.53}{M^{0.5}}$[/tex]

[tex]$M^{0.5}= 5.088$[/tex]

M = 25.88 g/mol

Answer:

ITS ANSWER IS

OPTION B. ATOMIC NUMBER

HI HAVE A NICE DAY

Answer:

Na2S(aq)+Cd(No3)2(aq)=CdS(s)+2NaNo3(aq)

Answer: it’s checkbox 2&3

Stoichiometry refers to the relationship between the moles of reactants and products.

This question must be solved using both stoichiometry and the gas laws

The reaction equation is;

2 Al(s) + 6 HCl(aq) --------> 2 AICI3, (aq) + 3 H2(g)

Number of moles of Al = 3g/27g/mol = 0.11 moles

According to the reaction equation;

2 moles of Al yields 3 moles of H2

0.11 moles of Al yields 0.11 * 3/2 = 0.165 moles

From the ideal gas equation;

PV=nRT

P = ?

n= 0.165 moles

V = 0.85 L

T = 297 K

R = 0.082 atmLK-1mol-1

P= nRT/V

P = 0.165 * 0.082 * 297/0.85

P= 4.73 atm

P1 = 4.73 atm

T1 = 297 K

V1 = 0.85 L

P2 = 1 atm

T2 =273 K

V2 =?

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 =  4.73 * 0.85 * 273/1 * 297

V2 = 3.69 L

P1 = 760 mmHg

T1 = 273 K

V1 = 3.69

P2 = 438 mm Hg

T2 = 284.2 K

V2 =?

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 760 * 3.69 * 284.2/438 *273

V2 = 797010.48/119574

V2= 6.67 L

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Answer:

1176 grams

Explanation:

nH2SO4 =2*6=12 mol

mH2SO4=12*98=1176 grams

Answer:

solution given:

molarity of H₂SO₄=6 M

volume=2L

no of mole =6M*2=12mole

we have

mass =mole* actual mass=12*98=1176g

the mass is 1176g.

The question is incomplete. The complete question is :

Hydrogen [tex](H_2)[/tex] gas and oxygen [tex](O_2)[/tex] gas react to form water vapor [tex](H_2O)[/tex]. Suppose you have 11.0 mol of [tex]H_2[/tex] and 13.0 mol of [tex]O_2[/tex] in a reactor. Calculate the largest amount of [tex]H_2O[/tex] that could be produced. Round your answer to the nearest 0.1 mol .

Solution :

The balanced reaction for reaction is :

[tex]$2H_2(g) \ \ \ \ + \ \ \ \ \ O_2(g)\ \ \ \rightarrow \ \ \ \ 2H_2O(g)$[/tex]

11.0                      13.0

11/2                       13/1     (dividing by the co-efficient)

6.5 mol               13 mol    (minimum is limiting reagent as it is completely consumed during the reaction)

Therefore, [tex]H_2[/tex] is limiting reagent. It's stoichiometry decides the product formation amount from equation above it is clear that number of moles for [tex]H_2O[/tex] will be produced = number of moles of [tex]H_2[/tex]

                                     = 11.0 mol

Answer:

Transition temperature is the temperature at which a substance changes from one state to another.

Allotropy is the existence of an element in many forms.

Answer:

liquid will be evaporated while solid remains

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[tex]SilentNature[/tex]

The volume occupied by the given amount of sulfur dioxide will be 84 L.

please show working my dear citizen

Answer:

Explanation:

The given balanced chemical equation is:

[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]

The value of Kc at 445oC is 0.020.

[HI]=1.5M

[H2]=2.50M

[I2]=0.05M

The value of Qc(reaction quotient ) is calculated as shown below:

Qc has the same expression as the equilibrium constant.

[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]

Qc>Kc,

Hence, the backward reaction is favored and the formation of Hi is favored.

Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.

Answer: 8 for both

Explanation:

After the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.

First, we will write the balanced equation for the reaction

H₂SO₄ + BaCl₂  → BaSO₄ + 2HCl

This means 1 mole of BaCl₂ is needed to react completely with 1 mole of H₂SO₄ to give 1 mole of BaSO₄ and 2 moles of HCl

From the question, 50.0g of sulfuric acid is mixed with 40.0 grams of barium chloride. To determine the quantity of each substance remaining after the complete reaction, we will first determine the number of moles present in each of the reactant.

For H₂SO₄

mass = 50.0g

Molar mass = 98.079 g/mol

From the formula

Number of moles = Mass / Molar mass

∴ Number of moles of H₂SO₄ = 50.0g / 98.079 g/mol

Number of moles of H₂SO₄ = 0.5098 mol

For BaCl₂

mass = 40.0 g

Molar mass = 208.23 g/mol

∴ Number of moles of BaCl₂ = 40.0g / 208.23 g/mol

Number of moles of BaCl₂ = 0.1921 mol

Since the number of moles of H₂SO₄ is more than that of BaCl₂, then H₂SO₄ is the excess reagent and BaCl₂ is the limiting reagent (that is, it will be used up completely during the reaction)

From the equation, 1 mole of H₂SO₄ is needed to completely react with 1 mole of BaCl₂

∴ 0.1921 mol of H₂SO₄ will be needed to completely react with 0.1921 mol of BaCl₂.

Therefore, after the reaction is complete, 0 mole (i.e 0 grams) of BaCl₂ will remain and (0.5098 mole - 0.1921 mole) of H₂SO₄ will remain.

Number of moles H₂SO₄ that will remain = 0.5098 mole - 0.1921 mole = 0.3177 moles

Now, we will convert this to grams

From the formula

Mass = Number of moles × Molar mass

Mass of H₂SO₄ that will remain = 0.3177 moles × 98.079 g/mol

Mass of H₂SO₄ that will remain = 31.1597 g

Mass of H₂SO₄ that will remain ≅ 31.16 g

Hence, after the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.

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Answer:

The phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.

Explanation:

The structure of nucleic acid polymers is built up from monomers of nucleotides.

A nucleotide consists of a sugar backbone which is either a ribose or deoxyribose sugar, a nitogenous base which is either a purine or pyrimidine, and a phosphate group. The nitrogenous base is attached to the carbon number 1 or C-1 of the sugar backbone by a covalent bond. The phosphate group on the other hand is covalently attached to the carbon number 5 or 5' carbon of the sugar backbone.

When polymers of nucleic acids are formed, the phosphate at the 5' carbon of the sugar backbone is covalently linked in a phosphodiester bond to the 3' carbon of the sugar backbone in another nucleotide molecule, thus extending the strands of the nucleic acid molecule.

Nucleases are enzymes that break down the phosphodiseter bonds in nucleic acids resulting in nucleotide monomers. After complete digestion ofmthe nucleic acid polymer by nucleases, the phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.

Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.

At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.

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Answer:

Part A

HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)

Part B

ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)

Explanation:

The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;

RCOOH + NaOH ----> RCOONa + H2O

We have to note the fact that the net ionic reaction still remains;

H^+(aq) + OH^-(aq) ---> H2O(l)

In both cases, the reaction can occur and they actually do occur as written.

Answer:

Pure water has an acidity of about 7 on the pH scale. -is a chemical property of pure water. Pure water has an acidity of about 7 on the pH scale

Answer: không màu , không mùi không vị

Explanation:

Answer:

gggggggggg

Explanation:

gggggggg

The tripeptide formed from threonine, glycine and alanine is neutral at the pH of 7.3. The carboxylic end is negative charged by donating its proton to form the NH₃⁺ group.

Peptides are protein units formed from two or more amino acids bonded through peptide bonds. There are essential and non-essential amino acids. Essential amino acids have to be uptake from food and non-essential amino acids are synthesized inside the body.

Threonine is an essential amino acid with a CH₃CHOH side group. Glycine has the simplest side group hydrogen and alanine has  CH₃ side chain. Both glycine and alanine are non-essential amino acids.

Each amino acids are represented with a three letter code or one letter symbol. Thus threonine is T,  G for glycine and A for alanine. At a pH of 7.3 the peptide formed from these amino-acids contains a negatively charged carboxylic end.

A positively charged amino end made by protonation from the acid group make the overall charge zero. The structure of the peptide is given in the uploaded image.

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Answer:

C

Explanation:

According to Newton's first law, an object in motion will stay in motion unless acted upon by an unbalanced force. The option which gives the same meaning is :

According to Newton's first law, an object in motion as long as the net forces acting on it are zero will stay in:

C. constant motion in a straight line

A constant motion in mechanics is a number that remains constant throughout the motion, effectively placing a restriction on the motion.

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Answer:

1. more

2. less

3. more

4. less

Explanation:

Explanation:

here's the answer to the question

Explanation:

The given chemical reaction is:

[tex]2 N_2 (g) + O_2(g) -> 2 N_20 (g) delta Hrxn= +163.2 kJ[/tex]

When two moles of nitrogen reacts with oxygen, it requires 163.2kJ of energy.

When 5.00 mol of nitrogen requires how much energy?

[tex]5.00 mol x \frac{163.2 kJ }{2 mol} \\=408 kJ[/tex]

Hence, the answer is 408 kJ of heat energy is required.

Answer:

"They established the laws of planetary motion"

Explanation:

Mr. Kepler was the astronomer who came up with the "Laws of Planetary Motion."