When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier than ordinary seawater) is evaporated, 181 grams of MgCl2 are recovered. What is the molarity of MgCl2 in the original sample?

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:

atomic number

Explanation:

The mass number (A) of an element is obtained by summing the number of neutrons and protons (nucleons) in the atom of that element. However, the proton number of an element is its atomic number (Z). Hence, it can be said that:

mass number = no. of neutrons + atomic number/number of protons.

According to this question, an isotope of oxygen is said to have a mass number of 17. If the neutron number of this isotope is given as 9, then this means that its ATOMIC NUMBER or no. of protons of that isotope is (17 - 9) = 8.

Answer:

atomic number

Explanation:

Answer:

very warm water consuving

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:

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:

%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%

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

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Explanation:

While the sun is an excellent source of energy, not all forms of life can utilize the sun's energy directly. This lesson describes how plants transform the sun's energy into potential energy stored in sugar, how living organisms utilize energy in sugar to perform work, and how the relationship between photosynthesis and cellular respiration is necessary for life.

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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:

THE ANSWER IS: copper(I) sulfide.

hope this helped <3

Explanation:

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: 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:

D

Explanation:

8oz=1cup

Xoz=1/4cup

cross multiply

Xoz=8×1/4=2

2oz=1/4cup

1oz=2T

2oz=x

cross multiply

×=2×2=4T

The cup of butter, number of tablespoons and ounces of butter are all in direct proportion with one another. The interpretation of the information provided is:

(b). The information provided gives the conversion factor 8oz/1 cup.

Given that:

[tex]1\ recipe = \frac{1}{4}\ cup[/tex]

[tex]1\ oz = 2T[/tex]

[tex]8\ oz = 1\ cup[/tex]

Multiply both sides of [tex]1\ oz = 2T[/tex] by 8

[tex]8 \times 1oz = 8 \times 2T[/tex]

[tex]8 oz = 16T[/tex]

Substitute 8oz for 1 cup in: [tex]1\ recipe = \frac{1}{4}\ cup[/tex]

[tex]1\ recipe = \frac{1}{4} \times 8oz[/tex]

Substitute [tex]8 oz = 16T[/tex]

[tex]1\ recipe = \frac{1}{4} \times 16T[/tex]

[tex]1\ recipe = 4T[/tex]

The interpretations are as follows:

Hence, (b) is correct

Read more about direct proportions at:

https://brainly.com/question/1266676

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.

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:

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:

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:

a

Explanation:

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:

The answer is consumers

Answer:

Consumers are the group of organisms that has the least biomass and energy.

Answer :

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:

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:

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:

B

Explanation:

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

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.