Come up with a definition for density

Answer:

c) B, C

Explanation:

NaOH(aq) + HBr(aq) -----> NaBr(aq) +H2O(l)

1) concentration of acid CA= 0.05 M

Concentration of base CB= 0.1 M

Volume of acid VA= 25.00ml

Volume of base VB= unknown

Number of moles of acid NA= 1

Number of moles of base NB= 1

CAVA/CBVB = NA/NB

CAVANB =CBVBNA

VB= CAVANB/CB NB

VB= 0.05 × 25 × 1/ 0.1 ×1

VB= 12.5 ML

2.

Answer:

To calculate the [OH-] in the solution we must first find the pOH

That's

pH + pOH = 14

pOH = 14 - pH

First to find the pH we use the formula

From the question

[H3O+]= 2.6 × 10^-8 M

pH = - log 2.6 × 10^-8

pH = 7.6

pH = 8

So we pOH is

pOH = 14 - 8 = 6

To find the [OH-] we use the formula

6 = - log [OH-]

Find antilog of both sides

The solution is slightly basic since it's pH is in the basic region and slightly above the neutral point 7

Hope this helps you

Answer:

[tex]\Huge \boxed{\mathrm{Kr}}[/tex]

Explanation:

Krypton is an element in the periodic table with an atomic number of 36.

The symbol for Krypton is Kr.

Answer:

KR.

Explanation:

Use the periodic table for reference:

Answer:

i hope it work

Explanation:

as

accurate reading range = reading ± uncertainty

so you have to say about accurate reading that its,lies in range

=(6389-436) →(6389+436)

=5953→6825

and  the correct number of significant would be 3

Answer:

[tex]\Delta _rG=-4.3\frac{kJ}{mol}[/tex]

Explanation:

Hello,

In this case, for the given dissociation reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):

[tex]\Delta _rH=\Delta _fH_{NH^{4+}}+\Delta _fH_{NO_3^-}-\Delta _fH_{NH_4NO_3}\\\\\Delta _rH=-132.5+(-205.0)-(-365.6)=28.1kJ/mol[/tex]

Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):

[tex]\Delta _rS=S_{NH^{4+}}+S_{NO_3^-}-S_{NH_4NO_3}\\\\\Delta _rS=113.4+146.4-151.1=108.7J/mol*K[/tex]

Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the given temperature (298.15 K), we finally obtain (two significant figures):

[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\\Delta _rG=28.1kJ/mol-(298.15 K)(108.7\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\\Delta _rG=-4.3\frac{kJ}{mol}[/tex]

Best regards.

Answer:

-434.14 kJ

Explanation:

Step 1: Write the balanced equation

4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)

Step 2: Calculate the standard free energy change (ΔG°r) for the reaction

We will use the following expression.

ΔG°r = 4 mol × ΔG°f(NO(g)) + 6 mol × ΔG°f(H₂O(g)) - 4 mol × ΔG°f(NH₃(g)) - 5 mol × ΔG°f(O₂(g))

ΔG°r = 4 mol × (86.55 kJ/mol) + 6 mol × (-228.57 kJ/mol) - 4 mol × (-16.45 kJ/mol) - 5 mol × (0 kJ/mol)

ΔG°r = -959.42 kJ

Step 3: Calculate the standard free energy change for 1.81 moles of NH₃

959.42 kJ are released per 4 moles of NH₃.

[tex]\frac{-959.42 kJ}{4mol} \times 1.81mol = -434.14 kJ[/tex]

Answer:

Explanation:

Answer in attached file .

Question is incomplete, the complete question is as follows:

A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?

A. Toxicity, because it can be observed by altering the state of the substance

B. Boiling point, because it can be observed by altering the state of the substance

C. Toxicity, because it can be observed by replacing the atoms of the substance with new atoms

D. Boiling point, because it can be observed by replacing the atoms of the substance with new atoms

Answer:

B.

Explanation:

A student can examine a substance without altering the bonds within the molecules by examining its boiling point.

The boiling point is the property of a substance, at which the substance changes its state, which is from solid to liquid, liquid to gas and others. So, examining the boiling point will alter the bonds within the molecules as the state of substance will change.

Hence, the correct answer is "B".

Answer:

D) 2.01 x 10⁻⁴ .

Explanation:

pH = 2.11

[ H⁺ ] = [tex]10^{-2.11}[/tex]  

Let the acid be HA

It will ionise as follows .

                                        HA       ⇄       H⁺       +        A⁻

in equilibrium                 .30               [tex]10^{-2.11}[/tex]          [tex]10^{-2.11}[/tex]         

Acid ionisation constant Ka  =   [tex]\frac{(10^{-2.11})^2}{0.3}[/tex]

= 2 x 10⁻⁴                

Answer:

D) 2.01 x 10⁻⁴ is correct!

Explanation:

I got it in class!

Hope this Helps!! :))

Answer:

examples of things which contain potassium are:

green vegetables

root vegetables

fruits

potassium chloride

potassium sulphate

Explanation:

if you need a specific answer please send the options

Answer:

C

Explanation:

The answer is the one with 20 protons, 20 neutrons, and 6-8-8-2 electrons.

Answer:

0.16 M

Explanation:

Data provided as per the question is below:-

Volume of [tex]HNO_3[/tex] = 0.22 L

The Volume of NaOH = 0.18 L

Morality of NaOH = 0.2

According to the given situation, the calculation of the concentration of the [tex]HNO_3[/tex] is shown below:-

For equivalence,

Number of the equivalent of [tex]HNO_3[/tex] = Number of equivalents of NaOH

[tex]= \frac{0.18\times0.2}{0.22}[/tex]

[tex]= \frac{0.036}{0.22}[/tex]

= 0.16363 M

or

= 0.16 M

Answer:

Weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.

Explanation:

The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation:

[tex] pH = pKa + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]  

We have that pH = 10.3 and the Ka is 4.7x10⁻¹¹, so:

[tex] 10.3 = -log(4.7 \cdot 10^{-11}) + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]  

[tex] \frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]  (1)

Also, we know that:

[tex] [Na_{2}CO_{3}] + [NaHCO_{3}] = 0.20 M [/tex]    (2)

From equation (2) we have:

[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] [/tex]   (3)

By entering (3) into (1):

[tex] \frac{0.20 - [NaHCO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]

[tex] 0.94*[NaHCO_{3}] + [NaHCO_{3}] = 0.20 [/tex]

[tex] [NaHCO_{3}] = 0.103 M [/tex]  

Hence, the [Na_{2}CO_{3}] is:

[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] = 0.20 M - 0.103 M = 0.097 M [/tex]  

Now, having the concentrations and knowing the volume of the buffer solution we can find the mass of the sodium carbonate and the sodium bicarbonate, as follows:

[tex]m_{Na_{2}CO_{3}} = C*V*M = 0.097 mol/L*0.050 L*105.99 g/mol = 0.5141 g[/tex]

[tex]m_{NaHCO_{3}} = C*V*M = 0.103 mol/L*0.050 L*84.007 g/mol = 0.4326 g[/tex]

Therefore, to prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3 we need to weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.      

I hope it helps you!

Answer:

See explanation

Explanation:

We must remember that the D / L nomenclature refers to the orientation of the hydroxyl group on carbon 5. If the "OH" is on the right we will have a D configuration. Yes, the "OH" is on the left we will have an L configuration. (See figure 1)

Now, the orientation of this "OH" has nothing to do with the ability of the molecule to deflect polarized light. If the molecule deflects light to the left we will have the symbol "(-)" (levorotation) if the molecule deflects light to the right we will have the symbol "(+)" (dextrorotation).

So in the "D" configuration, we can have both a right (+) and a left (-) deviation.

I hope it helps!

Answer:

[tex]\% m/m= 14.3\%[/tex]

Explanation:

Hello,

In this case, the by mass percent is computed as shown below:

[tex]\% m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%[/tex]

Whereas the solute is the sugar and the solvent the water, therefore, the concentration results:

[tex]\% m/m=\frac{63g}{63g+378g} *100\%\\\\\% m/m= 14.3\%[/tex]

Best regards.

Answer:

8.90 M

Explanation:

Step 1: Given data

Step 2: Calculate the initial concentration

We have a concentrated NaCl solution and we want to prepare a diluted one. We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

C₁ = C₂ × V₂  / V₁

C₁ = 5.00 M × 1.13 L  / 0.635 L

C₁ = 8.90 M

Answer:

[tex]\large \boxed{\text{8.90 mol/L}}[/tex]

Explanation:

We can use the dilution formula to calculate the concentration of the original solution.

[tex]\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\\text{635 mL }\times c_{1} & = & \text{1130 mL} \times \text{5.00 mol/L}\\635 c_{1}&=& \text{5650 mol/L}\\c_{1}& = & \dfrac{5650}{635}\text{ mol/L}\\\\& = & \textbf{8.90 mol/L}\\\end{array}\\\text{The initial concentration was $\large \boxed{\textbf{8.90 mol/L }}$}[/tex]

Answer: D - 126.4g

Explanation:  

% Yield = Actual Yield/Theoretical Yield

38% = Actual Yield/332.5

38/100 = Actual Yield/332.5

(.38)(332.5) = 126.35 g = 126.4 g Actual Yield

Answer:

is D. the correct answer

Explanation:

I'm not sure if it is. Please let me know if I'm mistaking.

Answer:

A. The atomic mass is 243 amu, and the atomic number is 95.

The total kinetic energy of a body is known as Thermal energy. Option A

Thermal energy is the direct sum of all the available random kinetic energies of molecules.

Also note that thermal energy is directly proportional to temperature in Kelvin.

Thus, the total kinetic energy of a body is known as Thermal energy. Option A

Learn more about thermal energy here:

https://brainly.com/question/19666326

#SPJ1

Answer:

A.) Thermal energy

Explanation:

I got it correct on founders edtell

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are in equilibrium

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are the actual concentrations

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

Where X is defined as the reaction coordinate

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

[CrO₄²⁻] = 0.135M

Answer: Here are the complete options.

A spontaneous process is one that occurs very quickly. A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium. A spontaneous process is one that occurs without continuous input of energy from outside the system. A process is spontaneous if it must be continuously forced or driven.

The correct option is

A spontaneous process is one that occurs without continuous input of energy from outside the system.

A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium

Explanation:

spontaneous process is one that occurs without continuous input of energy from outside the system and occur on its own because spontaneous processes are thermodynamically favorable characterized by a decrease in the system's free energy, they do not need to be driven by an outside source of energy. Which means that the initial energy is higher than the final energy.

A process that is spontaneous in one direction is nonspontaneous in the other direction under a given set of conditions, provided the system is not at equilibrium which will result to The sign of ΔG will change from positive to negative (or vice versa) where T = ΔH/ΔS. In cases where ΔG is: negative

Answer:

A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration  

B) The volume of NaOH required to permanently change the indicator at the end point is  a  drop of NaOH

c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same

D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same

Explanation:

A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration  

B) The volume of NaOH required to permanently change the indicator at the end point is  a  drop of NaOH

c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same

D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same

Answer:

water molecules

Explanation:

Redox reactions are carried out under acidic or basic conditions as the case may be.

If the reaction is carried out in an acid medium, then we must balance the hydrogen ions on the lefthand side of the reaction equation with water molecules on the righthand side of the reaction equation.

For instance, the equation for reduction of MnO4^- under acidic condition is shown below;

MnO4^-(aq) + 5e + 8H^+(aq) --------> Mn^2+(aq) + 4H2O(l)

Answer:

rate = [NO]²[H₂]

Explanation:

2NO + H2 ⟶N2 + H2O2 (slow)

H2O2 + H2 ⟶2H2O (fast)

From the question, we are given two equations.

In chemical kinetics; that is the study of rate reactions and changes in concentration. The rate law is obtained from the slowest reaction.

This means that our focus would be on the slow reaction. Generally the rate law is obtained from the concentrations of reactants in a reaction.

This means our rate law is;

rate = [NO]²[H₂]

Answer:

The correct answer is 1.60 Liters.

Explanation:

The given reaction:

CO (g) + H₂(g) ⇔ CH₃OH (g)

Based on the given reaction, two moles of H₂ reacts with one mole of CO and produce one mole of CH₃OH.

It is mentioned that 3.20 L of H₂ is reacted, therefore, there is a need to convert it into moles.

As 22.4 L at standard temperature and pressure is equivalent to 1 mole.

Therefore, 1 L at STP will be, 1/22.4 mole

Now 3.20 L at STP will be,

= 1/22.4 × 3.20

= 0.1428 mole

And as mentioned in the reaction that 2 moles of H₂ gives 1 mole of CH₃OH, therefore, 1 mole of H₂ will give 1/2 mole of CH₃OH

Now, 0.1428 mole of H₂ will give,

= 0.1428/2 = 0.071 mole of CH₃OH

= 0.071 × 22.4 = 1.60 L

The volume, in liters, of CH₃OH gas formed is 1.60 L

From the question,

We are to determine the volume of CH₃OH formed

The given chemical equation for the reaction is

CO(g)+ H₂(g) → CH₃OH

The balanced chemical equation for the reaction is

CO(g)+ 2H₂(g) → CH₃OH

This means

1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH

Now, we will determine the number of moles of H₂ present in the 3.20 L H₂ at STP

1 mol of an ideal gas has a volume of 22.4 L at STP

Then,

x mole of the H₂ gas will have a volume of 3.20 L at STP

x = [tex]\frac{3.20 \times 1}{22.4}[/tex]

x = 0.142857 mole

∴ The number of mole of H₂ present is 0.142857 mole

Since

2 moles of H₂ reacts to produce 1 mole of CH₃OH

Then,

0.142857 mole of H₂ will react to produce [tex]\frac{0.142857}{2}[/tex] mole of CH₃OH

[tex]\frac{0.142857}{2} = 0.0714285[/tex]

∴ The number of moles of CH₃OH produced = 0.0714285 mole

Now, for the volume of CH₃OH formed

Since

1 mol of an ideal gas has a volume of 22.4 L at STP

Then,

0.0714285 mol of CH₃OH will have a volume of 22.4 × 0.0714285  at STP

22.4 × 0.0714285 = 1.5999984 L ≅ 1.60 L

Hence, the volume of CH₃OH gas formed is 1.60 L

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Answer:

For the mentioned reaction, the balanced chemical equation is:  

KBr (aq) + AgNO3 (s) ⇒ KNO3 (aq) + AgBr (s)

The number written in front of the ion, atoms, and molecules in a chemical reaction so that each of the elements on both the sides of reactants and products of the equation gets balanced is known as the stoichiometric coefficient.  

From the mentioned balanced equation, the stoichiometric coefficient before KBr is 1, AgNO3 is 1, KNO3 is 1, as well as before AgBr is also 1. Thus, it is clear that 1 mole of potassium bromide reacts with 1 mole of silver nitrate to produce 1 mole of potassium nitrate and 1 mole of silver bromide.  

Answer:

30.12 mL.

Explanation:

We'll begin by calculating the molarity of the phosphoric acid. This can be obtained as follow:

Phosphoric acid H3PO4 will dissociate in water as follow:

H3PO4(aq) <==> 3H^2+(aq) + PO4^3-(aq)

From the balanced equation above,

1 mole of H3PO4 produces 3 moles of H+.

Therefore, XM H3PO4 will produce 8.70 M H+ i.e

XM H3PO4 = 8.70/3

XM H3PO4 = 2.9 M.

Therefore, the molarity of the acid solution, H3PO4 is 2.9 M.

Next, we shall write the balanced equation for the reaction. This is illustrated below:

2H3PO4 + 3Ba(OH)2 —> Ba3(PO4)2 + 6H2O

From the balanced equation above, we obtained the following:

Mole ratio of the acid, H3PO4 (nA) = 2

Mole ratio of the base, Ba(OH)2 (nB) = 3

Data obtained from the question include the following:

Molarity of base, Ba(OH)2 (Mb) = 6.50 M

Volume of base, Ba(OH)2 (Vb) =.?

Molarity of acid, H3PO4 (Ma) = 2.9 M

Volume of acid, H3PO4 (Va) = 45 mL

The volume of the base, Ba(OH)2 Needed for the reaction can be obtained as follow:

MaVa /MbVb = nA/nB

2.9 x 45 / 6.5 x Vb = 2/3

Cross multiply

2 x 6.5 x Vb = 2.9 x 45 x 3

Divide both side by 2 x 6.5

Vb = (2.9 x 45 x 3) /(2 x 6.5)

Vb = 30.12 mL

Therefore, the volume of the base, Ba(OH)2 needed for the reaction is 30.12 mL

Answer:

pH = 8.72

Explanation:

This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

As this compound acts like a base, we propose this equilibrium:

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰

Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M

So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

We can avoid the quadractic equation because Kb is so small

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH = 8.72

The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.

Since the following equation should be used.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

Now

(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

So,

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

Now

Kw = Ka. Kb

Kb = Kw/Ka

And,

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵

= 5.55×10⁻¹⁰

Now

[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

Now

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH

= 8.72

Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.

Learn more about an acid here: https://brainly.com/question/4519963

Answer:

1229.08975 kPa

Explanation:

Given that:

The depth of the water = 100 m below the surface of the water.

The density of the water = 1150 kgm

The mass of the water = depth of the water × density of the water

The mass of the water = 100 m × 1150 kgm

The mass of the water = 115000 kgf/m²

The mass of the water is the pressure of the water at a depth of 100 m below the surface of the water.

Since 1 kgf/m² = 0.00980665 kPa

Then  115000 kgm² = (115000  ×  0.00980665) kPa

=  1127.76475 kPa

At standard temperature and pressure , the atmospheric pressure = 101.325 kPa.

Therefore, the pressure below the surface of sea water  = 1127.76475 kPa +   101.325 kPa

= 1229.08975 kPa

Answer:

D. The Ca[OH]2 solution may have been unsaturated

Explanation:

The solubility product constant Ksp of any given chemical compound is a term used to describe the equilibrium between a solid and the ions it contains solution. The value of the Ksp indicates the extent to which any compound can dissociate into ions in water. A higher the Ksp, implies more greater solubility of the compound in water.

If the Ksp is more than the value in literature, this false value must have arisen from the fact that the solution was unsaturated hence it appears to be more soluble than it should normally be when saturated.