complete and balanced the following equations by predicting the products of each reaction pair.
NA + LiCI -> _ + _

Answer :

Answer:

0.124 mol Ag

General Formulas and Concepts:

Atomic Structure

Stoichiometry

Explanation:

Step 1: Define

[Given] 7.49 × 10²² atoms Ag

[Solve] mol Ag

Step 2: Identify Conversions

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

0.124377 mol Ag ≈ 0.124 mol Ag

Topic: AP Chemistry

Answer:

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

3 NO₂ + H₂O → 2 HNO₃ + NO

Explanation:

We will balance the equation using the trial and error method.

C₂H₅OH + O₂ → CO₂ + H₂O

1) We balance C atoms by multiplying CO₂ by 2 and H atoms by multiplying H₂O by 3.

C₂H₅OH + O₂ → 2 CO₂ + 3 H₂O

2) We balance O atoms by multiplying O₂ by 3.

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

NH₃ + O₂ → NO₂ + H₂O

1) We balance H atoms by multiplying NH₃ by 2 and H₂O by 3.

2 NH₃ + O₂ → NO₂ + 3 H₂O

2) We balance N atoms by multiplying NO₂ by 2.

2 NH₃ + O₂ → 2 NO₂ + 3 H₂O

3) We balance O atoms by multiplying O₂ by 3.5

2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O

C₃H₈ + O₂ → CO₂ + H₂O

1) We balance C atoms by multiplying CO₂ by 3 and H atoms by multiplying H₂O by 4.

C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

2) We balance O atoms by multiplying O₂ by 5.

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

H₂SO₄ + NaOH → Na₂SO₄ + H₂O

1) We balance Na atoms by multiplying NaOH by 2.

H₂SO₄ + 2 NaOH → Na₂SO₄ + H₂O

2) We balance H and O atoms by multiplying H₂O by 2.

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

NO₂ + H₂O → HNO₃ + NO

1) We balance H atoms by multiplying HNO₃ by 2.

NO₂ + H₂O → 2 HNO₃ + NO

2) We balance N atoms by multiplying NO₂ by 3.

3 NO₂ + H₂O → 2 HNO₃ + NO

Fe có tác dụng với Hcl

Fe + 2hcl -> fecl2 +h2

Answer:

có

Explanatio:

Answer: The correct answer is Protein B.

Explanation:

Gel-filtration chromatography is a separation technique that is based on the size of the molecules in a compound. It is also known as size-exclusion chromatography in which the eluent (carrier) used is an aqueous solution.

The matrix that is used is a porous material. When the sample is inserted in the column, the smaller particles interact strongly with the matrix than the large ones. Thus, as the eluent is passed through the matrix, larger molecules come out first, and the smallest molecule comes out last.

Given sizes of the proteins:

Protein A: 1200 kDa

Protein B: 2000 kDa

Protein C: 800 kDa

As protein B has the largest size of all the given proteins, it will emerge out first from the column.

Hence, the correct answer is Protein B.

Answer:

THE ANSWER IS: copper(I) sulfide.

hope this helped <3

Explanation:

Answer:

A) Molecular weight

B) Rate of diffusion

C) Agar plates

Time of diffusion

Explanation:

A) The independent variable is the molecular weight because it is the variable that the researcher changes at will to aid his research. It doesn't depend on other variables.

B) The dependent variable is "Rate of diffusion because it depends on the molecular weight of the compound.

C) A control variable is one that would be held constant throughout the research.

In this case, the agar plates and the time of 2 hours diffusion remain the same throughout.

Answer:

very warm water consuving

Answer:

Compare the solubility of silver iodide in each of the following aqueous solutions:

a. 0.10 M AgCH3COO

b. 0.10 M NaI

c. 0.10 M KCH3COO

d. 0.10 M NH4NO3

1. More soluble than in pure water.

2. Similar solubility as in pure water.

3. Less soluble than in pure water.

Explanation:

This can be explained based on common ion effect.

According to common ion effect the solubility of a sparingly soluble salt decreases further in a solution which has a common ion to it.

The solubility of AgI(s) silver iodide in water is shown below:

[tex]AgI(s) <=> Ag^{+}(aq)+I^{-}(aq)\\[/tex]

a. a. 0.10 M AgCH3COO has a common ion Ag+ with AgI.

So, AgI is less soluble than in pure water in this solution.

b. 0.10 M NaI has a common ion I- with AgI.

So, AgI is less soluble than in pure water in this solution.

c. 0.10 M KCH3COO:

This solution has no common ion with AgI.

So, AgI has similar solubility as in pure water.

d. 0.10 M NH4NO3:

In this solution, AgI can be more soluble than in pure water.

Answer:

Eu(ClO3)3

Explanation:

The chlorate ion is written as follows, ClO⁻ ₃. We can see from this that the ion is univalent.

From the formula, Eu203, it is easy to see that the europium ion is trivalent.

Hence, when a compound is formed between the europium ion and chlorate ion, the compound will be written as Eu(ClO3)3.

This is so because, when ionic compounds are formed, there is an exchange of valence between the ions in the compound. This gives the final formula of the ionic substance.

Answer:

1.46 × 10⁻⁸ g

Explanation:

Step 1: Given data

Molecules of CO₂: 2.00 × 10¹⁴ molecules

Step 2: Convert molecules to moles

We need a conversion factor: Avogadro's number. There are 6.02 × 10²³ molecules in 1 mole of molecules.

2.00 × 10¹⁴ molecules × 1 mol/6.02 × 10²³ = 3.32 × 10⁻¹⁰ mol

Step 3: Convert moles to mass

We need a conversion factor: the molar mass. The molar mass of CO₂is 44.01 g/mol.

3.32 × 10⁻¹⁰ mol × 44.01 g/mol = 1.46 × 10⁻⁸ g

Answer:

[tex]k_1=0.107s^{-1} \\\\k_2=0.711M^{-1}s^{-1}[/tex]

Explanation:

Hello there!

In this case, according to the given information and the attached picture in which we can see the units of the rate constant, it turns out possible for us to realize the two called rate laws are:

[tex]r=k[A]\\\\r=k[A]^2[/tex]

The former is first-order and the latter second-order; in such a way, we solve for the rate constant in both cases to obtain the following:

[tex]k=\frac{r}{[A]}=\frac{1.6x10^{-2}M/s}{0.15M}=0.107s^{-1} \\\\k=\frac{r}{[A]^2}=\frac{1.6x10^{-2}M/s}{(0.15M)^2}=0.711M^{-1}s^{-1}[/tex]

Regards!

Answer:

No reaction occurs.

Explanation:

The molecular reaction is as follows;

MgSO4(aq) + Ni(NO3)2(aq) ----> Mg(NO3)2(aq) + NiSO4(aq)

We can see from the reaction above that the both products of the reaction are soluble. Recall that a double replacement reaction often yields one insoluble product which separates as a precipitate.

This reaction does not occur since the two products that ought to be obtained are soluble in water.

Answer:

Ammonium hydroxide, NH₄OH

Magnesium hydroxide, Mg(OH)₂

Sodium hydroxide, NaOH

Lithium hydroxide, LiOH

Explanation:

A base is a substance which neutralizes acids to produce salt and water. Bases are hydroxide or oxides of metals. Bases may be soluble or insoluble in water. Bases generally have a bitter taste and turn red litmus paper or indicator red.

Alkalis are bases which are soluble in water. They form the hydroxide of the alkali metals or alkaline earth metals in solution and they ionize to produce hydroxide ions. They are slippery to touch and turn red litmus blue being bases.

Therefore, all alkalis are bases but not all bases are alkalis. Insoluble bases are not alkalis.

From the given chemical compounds the alkalis present in the list are:

Ammonium hydroxide, NH₄OH; since it is soluble in water and produces hydroxide ions

Magnesium hydroxide, Mg(OH)₂; since it is slightly soluble in water and produces hydroxide ions

Sodium hydroxide, NaOH; since it is soluble in water and produces hydroxide ions

Lithium hydroxide, LiOH; since it is soluble in water and produces hydroxide ions

CuO(copper oxide) is a base but not an alkali as it does not produce hydroxide ions.

Zn(OH)2 (zinc hydroxide) is amphoteric and is insoluble

MgO(magnesium oxide) is a base but not an alkali as it does not produce hydroxide ions.

Na2O(sodium oxide) is a base but not an alkali as it does not produce hydroxide ions.

CoO(cobalt oxide) is a base but not an alkali as it does not produce hydroxide ions.

Answer:

c) most have high electronegativities

Explanation:

They tend to have high electric CONDUCTIVITY because of the free-flowing d-orbital electrons, but have low electron affinity, ionization energy, and electronegativities.

Answer:

propanal

Explanation:

hope this helps :)

Answer:

i) 16.5g of ZnO

ii) 9.8 dm³ of NO2

Explanation:

The working is shown in the photo so kindly refer to it

Answer:

Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.

Sulfur dioxide: SO₂.

Beryllium hydride: BeH₂

Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.

Copper(II) oxide: CuO.

Calcium sulfate: CaSO₄

Nitric acid: HNO₃.

Explanation:

Hello there!

In this case, it seems that the formulas were not given, however, we can write the correct one for each given compound according to the widely used nomenclature rules as shown below:

Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.

Sulfur dioxide: SO₂.

Beryllium hydride: BeH₂

Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.

Copper(II) oxide: CuO.

Calcium sulfate: CaSO₄

Nitric acid: HNO₃.

Regards!

Answer: The mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g

Explanation:

Molarity is calculated by using the equation:

[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)  

Molarity of [tex]H_2SO_4[/tex] = 3.0 M

Volume of solution = 25.0 L

Putting values in equation 1, we get:

[tex]\text{Moles of }H_2SO_4=(3.0mol/L\times 25.0L)=75mol[/tex]

The ideal gas equation is given as:

[tex]PV=nRT[/tex] .......(2)

where,

P = pressure of the gas = 0.68 atm

V = volume of gas = [tex]3.1\times 10^3L[/tex]

n = number of moles of gas = ? moles

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

T = temperature of the gas = 298 K

Putting values in equation 2, we get:

[tex]0.68atm\times 3.1\times 10^3L=n\times 0.0821L.atm/mol.K\times 298K\\\\n=\frac{0.68\times 3.1\times 10^3}{0.0821\times 298}=86.16mol[/tex]

For the given chemical equation:

[tex]NH_3(g)+H_2SO_4(aq)\rightarrow (NH_4)_2SO_4(aq)[/tex]

By stoichiometry of the reaction:

If 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]NH_3[/tex]

So, 75 moles of [tex]H_2SO_4[/tex] will react with = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex]NH_3[/tex]

As the given amount of [tex]NH_3[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent

Thus, [tex]H_2SO_4[/tex] is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 1 mole of [tex]H_2SO_4[/tex] produces 1 mole of [tex](NH_4)_2SO_4[/tex]

So, 75 moles of [tex]H_2SO_4[/tex] will produce = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex](NH_4)_2SO_4[/tex]

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We know, molar mass of [tex](NH_4)_2SO_4[/tex] = 132.14 g/mol

Putting values in above equation, we get:

[tex]\text{Mass of }(NH_4)_2SO_4=(75mol\times 132.14g/mol)=9910.5g[/tex]

Hence, the mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g

Answer:

%C = 60.9%; %H = 15.4%; %N = 23.7%

Explanation:

Step 1: Given data

Step 2: Calculate the percent composition of this compound

To calculate the percent by mass of any element (E), we will use the following expression.

%E = mE/m × 100%

%C = 7.34 g/12.04 g × 100% = 60.9%

%H = 1.85 g/12.04 g × 100% = 15.4%

%N = 2.85 g/12.04 g × 100% = 23.7%

Answer:

37.9% v/v

Explanation:

Since both the alcohol and solution are presumed to be liquid, this concentration can be expressed as a volume concentration (or % v/v):

volume concentration = volume of solute / volume of solution

[tex]\% v/v = 55/145= 0.379[/tex]

Answer:

the equilibrium concentration of HF is 2.85 M

Option a) 2.85 M is the correct answer.

Explanation:

Given the data in the question;

     H₂         +      F₂      ⇄     2HF

I    1.69 M        1.69 M           0

C    -x                 -x               +2x

E    1.69-x         1.69-x          +2x

given that Kc = 115        

Kc = [ HF ]² / [H₂][F₂]

we substitute

115 = [ 2x ]² / [ 1.69-x  ][ 1.69-x ]

lets find the square root of both sides

10.7238 = 2x / [ 1.69-x  ]

10.7238[ 1.69-x  ] = 2x

18.123222 - 10.7238x = 2x

2x + 10.7238x = 18.123222

12.7238x = 18.123222

x = 18.123222 / 12.7238

x = 1.424356

Hence, equilibrium concentration of HF = 2x

that is;

HF = 2 × 1.424356

HF = 2.8487 ≈ 2.85 M

Therefore, the equilibrium concentration of HF is 2.85 M

Option a) 2.85 M is the correct answer.

Answer:

B

Explanation:

Do diện tích tiếp xúc ở dạng bột cao hơn dạng tinh thể

The organic compounds that are capable of being a reactant in all substitution reactions would belong to the alkane group.

Alkanes are saturated compounds that ordinarily will not participate in addition reactions due to their saturated nature.

Thus, alkanes are only able to participate in substitution reactions involving the substitution of one or more of their component atoms for another atom.

This is unlike alkyne and alkenes which are naturally unsaturated. The unsaturation makes them a candidate for additional reactions.

More on saturation and substitution reactions can be found here: https://brainly.com/question/3735006

#SPJ2

Answer:

The theoretical yield potassium permanganate, KMnO₄ when starting with 50.0 g MnO₂ is 90.8 g

Explanation:

Molar mass of MnO₂ = (55 + 2 × 16) = 87.0 g/mol

Molar mass of KMnO₄ = (39 + 55 + 4 × 16) = 158 g/mol

Moles of MnO₂ in 50 g = reacting mass / molar mass

where reacting mass = 50 g

Moles of MnO₂ in 50 g = 50 g /87 g/mol = 0.575 moles

The equation for the production of potassium permanganate is as follows:

2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2

From the equation of the reaction above, 2 moles of MnO₂ produces 2 moles of KmNO₄. The mole ratio of MnO₂ to KMnO₄ is 1 : 1

Therefore, 0.575 moles of MnO₂ will produce theoretically 0.575 moles of KMnO₄

Mass of 0.575 moles of KMnO₄ = number of moles × molar mass

Mass of 0.575 moles of KMnO₄ = 0.575 moles × 158 g/mol = 90.8 g of KMnO₄

Therefore, the theoretical yield potassium permanganate when starting with 50.0 g MnO₂ is 90.8 g

Answer:

0.0270w/v% H3PO4 in the soda

Explanation:

All phosphates reacts producing Ag3PO4. To solve this question we must convert the mass of Ag3PO4 to moles. These moles = moles of H3PO4. We can find, thus, the mass of H3PO4 and the w/v% as follows:

Moles Ag3PO4 -Molar mass: 418.58g/mol-

0.0576g * (1mol / 418.58g) = 1.376x10⁻⁴ moles Ag3PO4 = moles H3PO4

Mass H3PO4 -Molar mass: 97.994g/mol-

1.376x10⁻⁴ moles Ag3PO4 = moles H3PO4 * (97.994g/mol) = 0.0135g H3PO4

w/v%:

0.0135g H3PO4 / 50.0mL * 100 =

Answer:

Option A. Solution B is basic, and solution A is neutral.

Explanation:

The pH of a solution is simply defined as the measure of acidity or alkalinity of a solution.

The pH scale ranges from 0 to 14 with the following readings:

0 to 6 => Acidic solution

7 => Neutral solution

8 to 14 => Alkaline / basic solution.

From the above, we understood that solutions with pH ranging from 0 to 6 are acidic solutions. Those with pH of 7 are neutral solutions while those with pH ranging from 8 to 14 are basic solutions.

With the above information in mind, let us answer the question given above. This is illustrated below:

pH of solution A = 7

pH of solution B = 14

Solution A has a pH of 7. This implies that solution A is a neutral solution

Solution B has a pH of 14. This implies that solution B is a basic solution.

Thus, option A gives the correct answer to the question.