Please help meeee?!!!!

Answer:

.04 cg/L

Explanation:

The question is incomplete, the complete question is;

Groups of the periodic table correspond to elements with a. the same color b. the same atomic number c. similar chemical properties d. similar numbers of neutrons

Answer:

similar chemical properties

Explanation:

In the periodic classification of elements, elements are divided into groups and periods. Elements in the same group of the periodic table have the same number of outermost electrons and share very similar chemical properties.

Elements in the same period have the same number of shells and the same maximum energy level of the outermost electron. Chemical properties carry markedly across a period.

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:

MoO₃

Explanation:

To solve this question we must find the moles of molybdenum in Mo2O3. The moles of Mo remain constant in the new oxide. With the differences in masses we can find the mass of oxygen and its moles obtaining the empirical formula as follows:

Moles Mo2O3 -Molar mass: 239,878g/mol-

11.79g * (1mol / 239.878g) = 0.04915 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.09830 moles Mo

Mass Mo in the oxides:

0.09830 moles Mo * (95.95g/mol) = 9.432g Mo

Mass oxygen in the new oxide:

14.151g - 9.432g = 4.719g oxygen

Moles Oxygen:

4.719g oxygen * (1mol/16g) = 0.2949 moles O

The ratio of moles of O/Mo:

0.2949molO / 0.09830mol Mo = 3

That means there are 3 moles of oxygen per mole of Molybdenum and the empirical formula is:

Answer:

0.113 M

Explanation:

The reaction that takes place is:

First we convert 0.3967 g of NaHCO₃ into moles, using its molar mass:

As 1 mol of NaHCO₃ reacts with 1 mol of HCl, in 41.77 mL of the HCl solution there were 4.72x10⁻³ moles of HCl.

With the calculated number of moles and the given volume we calculate the concentration of the solution:

Answer:

A

Explanation:

Answer:

hi I used your code you got it

Answer:

the correct answer is c

Explanation:

becuase i had the same question

Answer:

Mg will be the limiting reagent.

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Being the molar mass of each compound:

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

0.0350 g of Mg is reacted with 10.00 mL (equal to 0.01 L) of 6 M HCl.

Molarity being the number of moles of solute that are dissolved in a certain volume, expressed as:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

in units [tex]\frac{moles}{liter}[/tex]

then, the number of moles of HCl that react is:

[tex]6 M=\frac{number of moles of HCl}{0.01 L}[/tex]

number of moles of HCl= 6 M*0.01 L

number of moles of HCl= 0.06 moles

Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl react with 24.3 grams of Mg, 0.06 moles of HCl react with how much mass of Mg?

[tex]mass of Mg=\frac{0.06 moles of HCl* 24.3 grams of Mg}{2 moles of HCl}[/tex]

mass of Mg= 0.729 grams

But 0.729 grams of Mg are not available, 0.0350 grams are available. Since you have less mass than you need to react with 0.06 moles of HCl, Mg will be the limiting reagent.

The limiting reactant in the reaction is Magnesium (Mg)

From the question,

We are to determine the limiting reactant in the reaction.

The given balanced chemical equation for the reaction is

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

This means

1 mole of Mg is required to react completely with 2 moles of HCl

Now, we will determine the number of moles of each reactant present

Mass = 0.0350 g

Using the formula

[tex]Number\ of\ moles = \frac{Mass}{Atomic\ mass}[/tex]

Atomic mass of Mg = 24.305 g/mol

∴ Number of moles of Mg present = [tex]\frac{0.0350}{24.305}[/tex]

Number of moles of Mg present = 0.00144 mole

Concentration = 6M

Volume = 10.00 mL = 0.01 L

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles HCl present = 6 × 0.01

Number of moles HCl present = 0.06 mole

Since,

1 mole of Mg is required to react completely with 2 moles of HCl

Then

0.00144 mole of Mg is required to react completely with 2×0.00144 mole of HCl

2×0.00144 = 0.00288

∴ The number of moles of HCl required to react completely with the Mg is 0.00288 mole

Since the number of moles of HCl present is more than 0.00288 mole, then HCl is the excess reactant and Mg is the limiting reactant.

Hence, the limiting reactant in the reaction is Magnesium (Mg)

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Answer:

aasjajiakjka

Explanation:

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:

See explanation.

Explanation:

Hello there!

In this case, since this problem  is about gas laws, more specifically about the Gay-Lussac's one since the volume is said to be constant, we can use the following equation for its solution for the final pressure, P2:

[tex]\frac{P_2}{T_2} = \frac{P_1}{T_1}[/tex]

[tex]P_2= \frac{P_1T_2}{T_1}\\\\P_2 =\frac{12.0atm*450K}{300K}\\\\P_2= 18.0atm[/tex]

Thus, we fill in the table as follows:

                           Initial        Final

Pressure           12.0 atm     18.0 atm

Volume                4.0 L         4.0 L

Temperature      300K         450K

Regards!

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

Answer:

The most stable conformation of the following compound has

A. An axial methyl group and an axial ethyl group.

B. An axial methyl group and an equatorial ethyl group.

C. An axial tert-butyl group.

D. An equatorial methyl group and an equatorial ethyl group.

E. An equatorial methyl group and an axial ethyl group.

Explanation:

The most stable conformation in the cyclohexane ring is the one in which both the substituents are in the equatorial position.

Among the given options,

option D  An equatorial methyl group and an equatorial ethyl group.

When the substituents in the cyclohexane ring are in equatorial positions then, the steric repulsions will be reduced.

Answer is option D.

Answer:

Reactions involving the replacement of one atom or group of atoms. - Substitution reaction

Reactions involving removal of two atoms or groups from a molecule - Elimination reaction

Products show increased bond order between two adjacent atoms - Elimination reaction

Reactant requires presence of a π bond - Addition reaction

Product is the structural isomer of the reactant - Rearrangement reaction

Explanation:

When an atom or a group of atoms is replaced by another in a reaction, then such is a substitution reaction. A typical example is the halogenation of alkanes.

A reaction involving the removal of two atoms or groups from a molecule resulting in increased bond order of products is called an elimination reaction. A typical example of such is dehydrohalogenation of alkyl halides.

Any reaction that involves a pi bond is an addition reaction because a molecule is added across the pi bond. A typical example is hydrogenation of alkenes.

Rearrangement reactions yield isomers of a molecule. Rearrangement may involve alkyl or hydride shifts in molecules.

Reactions involving the replacement of one atom or group of atoms is substitution reaction, reactions involving removal of two atoms or groups from a molecule and products show increased bond order between two adjacent atoms is elimination reaction, reactant requires presence of a π bond in addition reaction and product is the structural isomer of the reactant is rearrangement reaction.

Chemical reactions are those reactions in which reactants undergoes through a variety of changes for the formation of new product.

Hence, classification of above points are done according to their characteristics.

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Answer:

i think C  is the answer

Explanation:

The change in average annual temperatures on earth will be due to "photosynthesis by plants and algae".

Photosynthesis can be defined as a process in which plants, as well as other organisms, as well as other organisms, utilize to transform sunlight into chemical energy which can then be released to power the organism's activities using cellular respiration.

Plants seem to be mostly photosynthetic eukaryotes belonging to the plantae kingdom.

Therefore, photosynthesis cannot change in average annual temperature on Earth.

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Answer: The correct option is C (One of the fatty acid salts was unsaturated, and it completely isomerized under the reaction conditions).

Explanation:

Fats and oils belongs to a general group of compounds known as lipids. Fatty acids are weak acid and are divided into two:

--> Saturated fatty acids: These have NO double bonds in their hydrocarbon chain, and

--> Unsaturated fatty acids: These have one or more double bonds in their hydrocarbon chain.

SAPONIFICATION is defined as the process by which fats and oil is hydrolyzed with caustic alkali to yield propane-1,2,3-triol and the corresponding sodium salt of the component fatty acids. During this process, One hydroxide ion is required to hydrolyze one ester linkage of a triacylglycerol molecule. Because there are three ester linkages in a triacylglycerol, three equivalents of sodium hydroxide will be needed to completely saponify the triacylglycerol. This explains the reason why saponification of the triacylglycerol iresulted in four different fatty acid salts.

Answer:

6.88 mg

Explanation:

Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄

The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.

175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P

Step 2: Calculate the rate constant for the decay of ³²P

The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.

k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹

Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days

For first-order kinetics, we will use the following expression.

ln P = ln P₀ - k × t

ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d

P = 6.88 mg

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:

since 1 mole = 6.022×10^23

hence in 2.32 moles no. of atoms= 6.022×10^23 × 2.32

=

13.97104 ×10^23

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Answer:

d

Explanation:

cuaase that it sirhal

The chemical formula PbCl₂(s) represented by:

A) a substance

The chemical formula PbCl₂(s) represents a substance. A substance is a single, pure chemical entity with a definite composition. It can be an element, a compound, or an alloy.

A solution is a homogeneous mixture of two or more substances. A homogeneous mixture is a mixture in which the components are evenly distributed throughout the mixture.

A heterogeneous mixture is a mixture in which the components are not evenly distributed throughout the mixture.

A reaction is a process in which one or more substances are transformed into one or more new substances.

Therefore, the chemical formula PbCl₂(s) represents a substance, and the answer is (A).

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Answer:

C

Explanation:

KE at B is max and PE is 0

KE at C is 0 and PE is max

so when student swings from C to B

its KE increases

and PE decreases

Answer:

a) 0.11

b)76.9

c) 8.8

d) 1.7*10^-4

Explanation:

Step 1: Data given

K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 2: Formula of K

aA(g) + bB(g) ⇌ cC(g) + dD(g)

K = [C]^c *[D]^d  / [A]^a * [B]^b

K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]

Step 3:

a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)

N2(g) + 3H2(g) ⇌ 2NH3

1/2N2 + 3/2H2(g) ⇌ NH3(g)    =>K' =  [tex]\sqrt{K}[/tex]

K' = [tex]\sqrt{1.3*10^-2}[/tex] = 0.11

b. 2NH3(g) ⇌ N2(g) + 3H2(g)

N2(g) + 3H2(g) ⇌ 2NH3

2NH3(g) ⇌ N2(g) + 3H2(g)    =>K' = 1/K

K' = 1/(1.3*10^-2) = 76.9

c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)

N2(g) + 3H2(g) ⇌ 2NH3

NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)    

=>K' = [tex]\frac{1}{\sqrt{K} }[/tex]

K' = [tex]\frac{1}{\sqrt{1.3*10^-2} }[/tex]

K' = 8.8

d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)

N2(g) + 3H2(g) ⇌ 2NH3

2N2(g) + 6H2(g) ⇌ 4NH3(g)

K' = K²

K' = (1.3*10^-2)²

K' = 1.7 *10 ^-4

Values of equilibrium constant at given temperature for the following reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.

Equilibrium constant is define as the ration of the concentrations of product to the concentrations of reactant with respect to the exponent of their coefficients.

Given chemical reaction is:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Equilibrium constant for this reaction is:

K = [NH₃]² / [N₂][H₂]³

K = 1.3 × 10⁻² (given)

1/2N₂(g) + 3/2H₂(g) ⇌ NH₃(g)

K₁ = √K

Because concentration of all given species is 1/2 of the given reaction, so value of K₁ will be written as:
K₁ = √(1.3 × 10⁻²) = 0.11

2NH₃(g) ⇌ N₂(g) + 3H₂(g)

K₂ = 1/K

Because concentration of reactant and products are reciprocal from the concentration of original given reaction, so value of K₂ will be written as:
K₂ = 1/1.3 × 10⁻² = 76.9

NH₃(g) ⇌ 1/2N₂(g) + 3/2H₂(g)

K₃ = 1/√K

Because concentrations of given species is reciprocal as well as half of the given original reaction, so value of K₃ will be written as:
K₃ = 1/√(1.3 × 10⁻²) = 8.8

2N₂(g) + 6H₂(g) ⇌ 4NH₃(g)

K₄ = K²

Because concentrations of given species is double of the given original reaction, so value of K₄ will be written as:
K₄ = (1.3 × 10⁻²)² = 1.7 × 10⁻⁴

Hence, the value of K for given reactions are 0.11, 76.9, 8.8 and 1.7 × 10⁻⁴ respectively.

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Answer:

Spontaneous at only low temperature. I made a chart for my AP Chem class if you want to refer to it.

Answer:

Magnesium reacts with dilute hydrochloric acid in a conical flask which is ... One student can add the magnesium ribbon to the acid and stopper the flask, ... 50 cm3 of 1M hydrochloric acid is a six-fold excess of acid.

Answer:

Explanation:

Continental tropical air masses are extremely hot and dry. Arctic, Antarctic, and polar air masses are cold. The qualities of arctic air are developed over ice and snow-covered ground. Arctic air is deeply cold, colder than polar air masses.

Answer:

0.01228s⁻¹ = rate constant

Half-life = 56.4s

Explanation:

The first order reaction follows the equation:

ln[A] = -kt + ln[A]₀

Where [A] is amount of reactant after time t = 45.0%, k is rate constante and [A]₀ initial amount of reactant = 100%

ln[45%] = -k*65s + ln[100%]

-0.7985 = -k*65s

0.01228s⁻¹ = rate constant

Half-life is:

Half-life = ln2 / k

The question is incomplete, the complete question is:

When 177. g of alanine [tex](C_3H_7NO_2)[/tex] are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is [tex]5.9^oC[/tex] lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is [tex]7.2^oC[/tex] lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

Answer: The van't Hoff factor for potassium bromide in X is 1.63

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

[tex]\Delta T_f=i\times K_f\times m[/tex]

OR

[tex]\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)

[tex]\Delta T_f=5.9^oC[/tex]

i = Vant Hoff factor = 1 (for non-electrolytes)

[tex]K_f[/tex] = freezing point depression constant

[tex]m_{solute}[/tex] = Given mass of solute (alanine) = 177. g

[tex]M_{solute}[/tex] = Molar mass of solute (alanine) = 89 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

[tex]5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m[/tex]

[tex]\Delta T_f=7.2^oC[/tex]

i = Vant Hoff factor = ?

[tex]K_f[/tex] = freezing point depression constant = [tex]2.37^oC/m[/tex]

[tex]m_{solute}[/tex] = Given mass of solute (KBr) = 177. g

[tex]M_{solute}[/tex] = Molar mass of solute (KBr) = 119 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

[tex]7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63[/tex]

Hence, the van't Hoff factor for potassium bromide in X is 1.63

Answer:

see explanation

Explanation:

SO₃(g) + 395.77 Kj/mole => S°(s) + 3/2O₂(g)

The standard heat of formation for SO₃(g) is given by the following rxn:

S°(s) + 3/2O₂(g) => SO₃(g) + 395.77 kJ/mole. Reversing this reaction is the decomposition of SO₃(g) into its basic elements in their standard state (25°C, 1atm) and is endothermic with +295.77Kj/mole.