At 1 atm, an unknown sample melts at 49.9 °C and boils at 209.5 °C. If the temperature is 0°C, what is the state of matter for the sample?

rate=0.2*3^1*3^2=0.2*3*9=5.4(mol/L)s so the correct answer is C.

Answer:

10.328 m

Explanation:

normal atmospheric pressure = 101325 Pa

density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3

pressure = pgh

where p = density

g = acceleration due to gravity = 9.81 m/s^2

h = height of column

imputing values, we have

101325 = 1000 x 9.81 x h

height of column h = 101325/9810 = 10.328 m

Answer:

1: 2-bromo-3-chloropentane

Explanation:

find longest carbon chain =5

place the Br and Cl on the carbon chain

follow naming rules I guess

Answer:

The correct answer is 0.00080 gram per liter.

Explanation:

Based on the given information, the solubility of water is 0.00380 gram per liter, the temperature mentioned is 25 degree C, the partial pressure of oxygen gas is 760 torr, and the mole fraction of oxygen is 0.210. There is a need to determine the solubility of oxygen in water.  

Based on Henry's law,  

Solubility of oxygen gas = Henry's constant × partial pressure of oxygen gas

Henry's constant, K = solubility of oxygen gas / partial pressure of oxygen gas

= 0.00380 g/L × 1 mol/32 grams / 760 torr × 1 atm/760 torr

= 0.00012 mol/L/atm

= 0.00012 M/atm

Now the partial pressure of the oxygen gas = mole fraction of oxygen × atmospheric pressure

= 0.210 × 1 atm

= 0.210 atm

Now putting the values in Henry's law equation we get,  

Solubility of oxygen gas = 0.00012 mol/L/atm × 0.210 at,

= 0.000025 mol/L × 32 gram/mol

= 0.00080 gram per liter

Answer:

pH = 3.39

Explanation:

The equilibrium in water of ascorbic acid (With its conjugate base) is:

H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)

Where the acidic dissociation constant is written as:

Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]

H₂O is not taken in the Ka expression because is a pure liquid.

As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:

[H₂C₆H₆O₆] = 2.5x10⁻³M - X

[HC₆H₆O₆⁻] = X

[H₃O⁺] = X

Replacing in the Ka expression:

7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]

1.975x10⁻⁷ - 7.9x10⁻⁵X = X²

0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷

Solving for X:

X = -0.00048566→  False solution, there is no negative concentrations

X = 0.00040666 → Right solution

As [H₃O⁺] = X, [H₃O⁺] = 0.00040666

pH is defined as -log [H₃O⁺];

pH = -log 0.00040666,

Answer:

7p

Explanation:

principal quantum number is 7

n=7( principle shell)

angular momentum quantum number gives sub shell

l = 1 means it is p orbital

so answer is 7p orbital

Answer:

B. Reactant + Reactant -> Product + Product

Explanation:

Reactants are substances that- as the name suggests- reacts with other substances at the beginning of a reaction

Products are substances that are produced as a result of the reaction

Typically, when writing a chemical reaction, an arrow is used to show the direction the reaction is moving.  In this case, the arrows in options B and C suggest that the reaction only moves in one direction- forwards

And as mentioned above, reactants are the substances at the start of the reaction, they're what mixes together to form a new product.  

To keep things simple:

Products can't be at the beginning of a reaction since they weren't formed yet.

Similarly, reactants can't be part of the products since they already existed and didn't need to be made. In a lot cases, the reactants would be completely used up to make the products

As such, only one possible chemical reaction would follow that reasoning:

    Reactant + Reactant ->  Product + Product

Reactant + Reactant → Product + Product is the correctly written chemical reaction. Hence, option B is correct.

A chemical equation is a mathematical expression of the chemical reaction which represents the product formation from the reactants.

In an equation, the reactants are written on the left-hand side and the products are written on the right-hand side demonstrated by one-headed or two-headed arrows.

Hence, option B is correct.

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Answer:

[tex]3Sr^{+2}+6Cl^{-}+6Li^{+}+2PO_{4}^{3-}-->Sr_{3}(PO_{4})_{2}+6Li^{+}+6Cl^{-}\\\\3Sr^{+2}+2PO_{4}^{3-} --->Sr_{3}(PO_{4})_{2}[/tex]

First equation is the complete ionic equation.

Second equation is the net ionic equation.

Answer:

[tex]14\\8[/tex]O

Explanation:

The top number always represents the mass number.

The bottom number always represents the atomic number.

The element always goes after the numbers.

If charge is present, that comes after the element.

The correct answer is B. on Apex!

They have been knocked away from atoms. Hence, option B is correct.

The photoelectric effect is a phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation.

When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect.

This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.

Hence, option B is correct.

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Answer:

The base is involved in the rate determining step of an E2 reaction mechanism

Explanation:

Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;

Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.

For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;

Rate = k[alkyl halide] [base]

The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.

Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.

Answer:

The correct answer is mass of Al3+ will be 3.23 grams and the mass of Co2+ will be 8.50 grams.

Explanation:

Based on the given information, 0.3731 L of 1.735 M of NaOH is added in a solution resulting in the precipitation of the ions as Al(OH)₃ and Co(OH)₂. Thus, the moles of NaOH will be molarity × V(L) = 1.735 × 0.3731 L = 0.647 moles.  

The mass of the precipitate given is 22.73 grams.  

Now let us assume that the mass of Al(OH)₃ will be x grams and the mass of Co(OH)₂ will be (22.73-x) grams

Therefore, the moles of Al(OH)₃ will be x grams/78 g/mol and as 3OH⁻ ions are needed so the moles will be 3x/78 mole.  

And, the moles of Co(OH)₂ will be (22.73-x)grams/92.94 g/mol and as 2OH⁻ ions are needed so the moles will be 45.46-2x/92.94 moles.

Now the equation will become,  

3x/78 + 45.46-2x/92.94 = 0.647 moles

0.03846 x + 0.489 - 0.02152 x = 0.647  

0.01694 x + 0.489 = 0.647

0.01694 x = 0.158

x = 0.158/0.01694

x = 9.327 grams

Hence, the mass of Al(OH)₃ is 9.327 grams, and the mass of Al³⁺ will be,  

= 9.327 gm/78 g/mol × 27 g/mol = 3.23 grams

Now the mass of Co(OH)₂ will be, (22.73 - 9.327) grams = 13.403 grams

the mass of Co²⁺ will be,  

= 13.403 grams / 92.94 g/mol × 58.94 g/mol = 8.50 grams

Answer:

Na₂CO₃  +  CaCl₂  →  CaCO₃  +  2 NaCl

Explanation:

Na₂CO₃  +  CaCl₂  →  CaCO₃  +  NaCl

First, determine what is balanced.  

Ca and CO₃ is balanced; there is one on each side.  

Na and Cl aren't balanced.  On one side, there is two of each, while on the other there is only one.  To fix this add a 2 in front of NaCl.  This will cause the equation to be balanced.

Na₂CO₃  +  CaCl₂  →  CaCO₃  +  2 NaCl

Answer:

Near the Hydrogen Atoms.

Explanation:

δ+ would be found near the dipole end of the hydrogens. δ- would be found near the dipole end of oxygen.

Answer:

3,4,1 and 6,5,2

Explanation:

In the electromagnetic spectrum the arrangement of the waves in increasing frequencies and decreasing wavelengths are as follows;

Radio waves

Microwaves

Infrared waves

Visible light rays

Ultraviolet rays

X-rays

Gamma rays

(a simple mnemonic is RMIVUXG)

Answer:

pH at equivalence point is 8.47

Explanation:

Benzoic acid react with NaOH, thus:

HC₇H₅O₂ + NaOH → C₇H₅O₂⁻ + H₂O + Na⁺

You reach equivalence point when moles of the acid = moles of NaOH.

Moles of benzoic acid are:

0.025L ₓ (0.0988mol / L) = 0.00247 moles

To have 0.00247 moles of NaOH in solution and reach equivalence point you need to add:

0.00247 moles NaOH ₓ (1L / 0.115mol) = 0.0215L of NaOH solution.

Total volume is 0.0465L.

There are produced 0.00247 moles of C₇H₅O₂⁻ and its molarity will be:

0.00247 mol C₇H₅O₂⁻ / 0.0465L = 0.0531M C₇H₅O₂⁻

C₇H₅O₂⁻ is in equilibrium with water, thus:

C₇H₅O₂⁻(aq) + H₂O ⇄ HC₇H₅O₂(aq) + OH⁻(aq)

Where Kb = Kw / Ka = 1x10⁻¹⁴ / 6.5x10⁻⁵ = 1.54x10⁻¹⁰ is:

Kb = 1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]

The concentrations in equilibrium of the species are:

[HC₇H₅O₂] = X

[OH⁻] = X

[C₇H₅O₂⁻] = 0.0531M - X

Where X represents how much C₇H₅O₂⁻ react, X is reaction coordinate

Replacing in Kb expression:

1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]

1.54x10⁻¹⁰ = [X] [X] / [0.0531 - X]

8.169x10⁻¹² - 1.54x10⁻¹⁰X = X²

8.169x10⁻¹² - 1.54x10⁻¹⁰X - X² = 0

Solving for X:

X = -2.858x10⁻⁶M → False solution, there is no negative concentrations

X = 2.858x10⁻⁶M → Right solution

As [OH⁻] = X

[OH⁻] = 2.858x10⁻⁶M

pOH is -log [OH⁻]

pOH = 5.54

pH = 14 - pOH

pH = 8.46

Answer:

A carboxylic acid group (-COOH) can form more hydrogen bonds with water than an ester group (-COO-).

Explanation:

The carboxylic acid group (-COOH) is found in the carboxylic acids. This group is ultimately responsible for the solubility of carboxylic acids in water. It is worthy of note that the high boiling points of low molecular weight carboxylic acids is often because they are capable of intermolecular hydrogen bonding which leads to the dimerization of carboxylic acid.

The solubility of carboxylic acids decreases as the length of the alkyl chain increases. Hence, a long chain carboxylic acid is less soluble in water than shorter chain carboxylic acids.

Ester molecules can't form hydrogen bonds with each other but they do form weak hydrogen bonds with water. This leads to the solubility of low molecular weight esters. However, if a carboxylic acid and an ester posses the same length of alky chain, the carboxylic acid will form more hydrogen bonds and thus be more soluble in water than than a corresponding ester of the same chain length.

Answer:

Option B

Explanation:

Scientific question are answered through experimentation, through testing the theory about the physical world.

Answer: its A

through observing and measuring the physical world

Explanation:

Answer:

WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.

Explanation:

Equation for the reaction:

K2SO4 + H20 ------->2 K+ + SO4^2-

When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.

1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-

At STP, 1 mole of K2SO4 will be the molar mass of the substance

Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol

Molar mass = 174 g/mol

So therefore;

1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion

When 63.7 g is used; we have:

174 g = 2 moles of K+

63.7 g = ( 63.7 * 2 / 174) moles of K+

= 0.73 moles of K+

Forr sulfate ion, we have:

174 g = 1 mole ofSO4^2-

63.7 g = (63.7 * 1 / 174) moles of SO4^2-

= 0.366 moles of SO4^2-

In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.

Answer:

A. [OH⁻] = 2.188x10⁻³M

B. pKb = 6.02

Explanation:

When hydrazine is in equilbrium with water, its reaction is:

N₂H₄(aq) + H₂O(l) ⇄ HN₂H₄⁺(aq) + OH⁻(aq)

Where Kb, is defined as the ratio between concentrations in equilibrium of the species, thus:

Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]

A. From pH, you can find [OH⁻], thus:

pH = -log [H⁺]

11.34 = -log [H⁺]

4.57x10⁻¹² = [H⁺]

As 1x10⁻¹⁴ = [OH⁻] [H⁺]

1x10⁻¹⁴ / 4.57x10⁻¹² = [OH⁻]

B. Concentrations in equilibrium of the species are:

[N₂H₄] = 5.0M - X

[HN₂H₄⁺] = X

[OH⁻] = X

Where X is reaction coordinate

As [OH⁻] = 2.188x10⁻³M

X = 2.188x10⁻³M

Replacing:

[N₂H₄] = 5.0M - 2.188x10⁻³M = 4.9978M

[HN₂H₄⁺] = 2.188x10⁻³M

[OH⁻] = 2.188x10⁻³M

Replacing in Kb expression:

Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]

Kb = [2.188x10⁻³M] [2.188x10⁻³M] / [4.9978M]

Kb = 9.577x10⁻⁷

pKb is defined as -log Kb

pKb = -log 9.577x10⁻⁷

Answer: That most of an atom's mass was packed in a central nucleus

Answer:

That most of an atom's mass was packed in the central nucleus.

Explanation:

Rutherford disproved Thomson's model in 1911 with his gold foil experiment, in which he demonstrated that the atom has a high-mass nucleus.

Answer:

The new solution is 1.4% m/V

Explanation:

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:

[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]

We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]

The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.

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Answer:

This reaction isn't yet at an equilibrium. It must shift in the direction of the reactant (namely [tex]\rm NOBr\; (g)[/tex]) in order to reach an equilibrium.

For this mixture, the reaction quotient is [tex]Q_c = 0.0126[/tex].

Explanation:

A reversible reaction is at equilibrium if and only if its reaction quotient [tex]Q_c[/tex] is equal to the equilibrium constant [tex]K_c[/tex].

Start by calculating the equilibrium quotient [tex]Q_c[/tex] of this reaction. Given the reaction:

[tex]\rm 2\; NOBr\; (g) \rightleftharpoons 2\; NO\; (g) + Br_2\; (g)[/tex].

Let [tex][\mathrm{NOBr\; (g)}][/tex], [tex][\mathrm{NO\; (g)}][/tex], and [tex][\mathrm{Br_2\; (g)}][/tex] denote the concentration of the three species. The formula for the reaction quotient of this system will be:

[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2}[/tex].

(Note, that in this formula, both [tex][\mathrm{NO\; (g)}][/tex] and [tex][\mathrm{NOBr\; (g)}][/tex] are raised to a power of two. That corresponds to the coefficients in the balanced reaction.)

Calculate the reaction quotient given the concentration of each species:

[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2} \approx 1.26\times 10^{-2} = 0.0126[/tex].

(Note that the unit is ignored.)

Apparently, [tex]Q_c > K_c[/tex]. Since [tex]Q_c[/tex] and [tex]K_c[/tex] are not equal, this reaction is not at an equilibrium. If external factors like temperature stays the same,

Keep in mind that [tex]Q_c[/tex] denotes a quotient. To reduce the value of a quotient, one may:

In [tex]Q_c[/tex], that means reducing the concentration of the products while increasing the concentration of the reactants. In other words, the system needs to shift in the direction of the reactants before it could reach an equilibrium.

Answer:

[tex]n_{Cl^-}=25molCl^-[/tex]

Explanation:

Hello,

In this case, since the given 5-M concentration of magnesium chloride is expressed as:

[tex]5\frac{molMgCl_2}{L}[/tex]

We can notice that one mole of salt contains two moles of chloride ions as the subscript of chlorine is two, in such a way, with the volume of solution we obtain the moles of chloride ions as shown below:

[tex]n_{Cl^-}=5\frac{molMgCl_2}{L}*\frac{2molCl^-}{1molMgCl_2} *2.5L\\\\n_{Cl^-}=25molCl^-[/tex]

Best regards.

Answer:

C. [tex]Cl^−_(_a_q_) + Ag^+_(_a_q_)->AgCl_(_s_)[/tex]

Explanation:

In this question our options are:

A. [tex]Ca^+^2_(_a_q_)+NO_3^-_(_a_q_)->Ca(NO_3)_2_(_a_q_)[/tex]

B. [tex]Ag^2^+_(_a_q_)+2Cl^-_(_a_q_)->AgCl_2_(_s_)[/tex]

C. [tex]Cl^−_(_a_q_) + Ag^+_(_a_q_)->AgCl_(_s_)[/tex]

D. None of the above because no reaction occurs

We have to remember that the ions produced by [tex]AgNO_3[/tex] are:

[tex]Ag^+[/tex]  and [tex]NO_3^-[/tex]

And the ions produced by [tex]CaCl_2[/tex] are:

[tex]Ca^+^2[/tex] and [tex]Cl^-[/tex]

Additionally, we will have a double displacement reaction so the compounds produce are:

[tex]AgCl[/tex] and [tex]Ca(NO_3)_2[/tex]

If we remember the solubility rules, all the nitrate salts are soluble and the salts made with silver are not soluble. With this in mind, we will have a solid-state for [tex]AgCl_(_s_)[/tex] and an aqueous state for  [tex]Ca(NO_3)_2_(_a_q_)[/tex].

If this is true, the final answer can be B or C. The charge of Ag is +1 so the final answer is C.

I hope it helps!

Answer:

550 m/s

Explanation:

The average molecular speed (v) is the speed associated with a group of molecules on average. We can calculate it using the following expression.

[tex]v = \sqrt{\frac{3 \times R \times T}{M} }[/tex]

where,

We can use the info of argon to calculate the temperature for both samples.

[tex]T = \frac{v^{2} \times M}{3 \times R} = \frac{(391m/s)^{2} \times 39.95g/mol}{3 \times 8.314J/k.mol} = 2.45 \times 10^{5} K[/tex]

Now, we can use the same expression to find the average molecular speed in a sample of Ne gas.

[tex]v = \sqrt{\frac{3 \times R \times T}{M} } = \sqrt{\frac{3 \times (8.314J/k.mol) \times 2.45 \times 10^{5}K }{20.18g/mol} } = 550 m/s[/tex]

Answer:

Option B. 0.136 g

Explanation:

The balanced equation for the reaction is given below:

2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)

Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:

Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol

Mass of AgNO3 from the balanced equation = 2 x 170 = 340g

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of Ag2O = (108x2) + 16 = 232g/mol

Mass of Ag2O from the balanced equation = 1 x 232 = 232g

Summary:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted with 80g of NaOH.

Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.

From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.

Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.

In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.

The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:

From the balanced equation above,

340g of AgNO3 reacted to produce 232g of Ag2O.

Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.

Therefore, 0.136g of Ag2O was produced from the reaction.