Write the balanced chemical equation for the reaction of the weak acid HCN with water. Include the phase of each species. How do you complete the Ka expression for this reaction?

Recall the heat capacity equation:

q = mc∆T

We're given mass, specific heat capacity, as well as the change in temperature. All we need to do is plug the numbers into the variables and we'll have our answer!

Although this question doesn't try to trick you, more often than not questions regarding energy change will attempt to throw you off with specific heat capacity. It's extremely important to note the units of the specific heat capacity and ensure that the numbers you use are in those units. As an example, the specific heat capacity might be given to you in J/mol*K - in this case, you'd have to do some unit conversions with your given data in order to fit all the numbers. In this question, we're given the specific heat capacity in J/gºC, so we don't need to change anything since all of our data is already in these units.

Anyways, back to the actual question:

q = mc∆T

q = (8.50 * [tex]10^{2}[/tex]) * (0.900) * (94.6 - 22.8)

q = 54927 (J)

Remeber to include significant figures:

54927 = 5.49 * 10^4 (J)

The required energy is 5.49 * [tex]10^{4}[/tex] Joules, or 5.49 * [tex]10^{1}[/tex] kJ

Let me know if you need any of my work to be explained!

- Breezeツ

Answer:

How to make scientific observations?

Observe something through your senses or record information using scientific tools and instruments and ask questions about your scientific observations (e.g., natural phenomena).

How to calculate data involving metric units?

To convert from one unit to another within the metric system usually means moving a decimal point (e.g., 1000000mm = 100000cm = 10000dm = 1000m = 100dkm = 10hm = 1km).

The balanced equation is:

(a) 2Mn^2+ + 2H2O2 + 4OH^- → 2MnO_2 + 4H2O.

(b)3SnO2^2- + 6OH^- + 2Bi(OH)3 → 3SnO3^2- + 2Bi + 9H2O. (c)14Cr2O7^2- + 7C2O4^2- + 22H2O → 4Cr^3+ + 14CO2 + 28H+ + 28e^-. (d)2ClO3^- + 16H^+ + 3Cl^- → 3Cl2 + 8H2O

(e)10BiO3^- + 60H^+ + 12Mn^2+ → 10Bi^3+ + 30H2O + 12MnO4^-

a. In the balanced redox equation Mn^2+ + H_2O_2 → MnO_2 + H_2O (in basic solution), the half-reactions are:

Reduction: Mn^2+ → MnO_2

Oxidation: H_2O_2 → H_2O

To balance the reduction half-reaction, we need to add four OH^- ions to the left side: Mn^2+ + 4OH^- → MnO_2 + 2H2O + 2e^-

To balance the oxidation half-reaction, we add four OH^- ions to the right side and water molecules to balance the oxygen atoms: H2O2 + 4OH^- → 2H2O + 2e^-

Now, multiply the reduction half-reaction by two and the oxidation half-reaction by one to equalize the electrons:

2(Mn^2+ + 4OH^- → MnO_2 + 2H2O + 2e^-)

H2O2 + 4OH^- → 2H2O + 2e^-

Finally, add the two half-reactions together and cancel out the common species: 2Mn^2+ + 2H2O2 + 4OH^- → 2MnO_2 + 4H2O

b. The balanced redox equation Bi(OH)3 + SnO2^2- → SnO3^2- + Bi (in basic solution) can be balanced by following these steps:

Reduction: SnO2^2- → SnO3^2-

Oxidation: Bi(OH)3 → Bi

To balance the reduction half-reaction, add two OH^- ions to the left side: SnO2^2- + 2OH^- → SnO3^2- + H2O + 2e^-

To balance the oxidation half-reaction, add six OH^- ions to the right side: Bi(OH)3 → Bi + 3H2O + 3e^-

Multiply the reduction half-reaction by three and the oxidation half-reaction by two to equalize the electrons:

3(SnO2^2- + 2OH^- → SnO3^2- + H2O + 2e^-)

2(Bi(OH)3 → Bi + 3H2O + 3e^-)

Combine the two half-reactions and cancel out the common species:

3SnO2^2- + 6OH^- + 2Bi(OH)3 → 3SnO3^2- + 2Bi + 9H2O

c. In the acidic solution, the balanced redox equation Cr2O7^2- + C2O4^2- → Cr^3+ + CO2 can be balanced as follows:

Reduction: Cr2O7^2- → Cr^3+

Oxidation: C2O4^2- → CO2

To balance the reduction half-reaction, add seven H2O molecules to the right side: Cr2O7^2- → 2Cr^3+ + 7H2O + 14e^-

To balance the oxidation half-reaction, add eight H^+ ions to the left side:

C2O4^2- + 2H2O → 2CO2 + 4H+ + 4e^-

Multiply the reduction half-reaction by two and the oxidation half-reaction by seven to equalize the electrons:

2(Cr2O7^2- → 2Cr^3+ + 7H2O + 14e^-)

7(C2O4^2- + 2H2O → 2CO2 + 4H+ + 4e^-)

Combine the two half-reactions and cancel out the common species:

14Cr2O7^2- + 7C2O4^2- + 22H2O → 4Cr^3+ + 14CO2 + 28H+ + 28e^-

d. In the acidic solution, the balanced redox equation ClO3^- + Cl^- → Cl2 + ClO2 can be balanced as follows:

Reduction: ClO3^- → ClO2

Oxidation: Cl^- → Cl2

To balance the reduction half-reaction, add two H^+ ions to the right side:

ClO3^- + 2H^+ → ClO2 + H2O + 2e^-

To balance the oxidation half-reaction, add two H2O molecules to the left side:

2Cl^- → Cl2 + 2e^-

Multiply the reduction half-reaction by two and the oxidation half-reaction by one to equalize the electrons:

2(ClO3^- + 2H^+ → ClO2 + H2O + 2e^-)

Cl^- → Cl2 + 2e^-

Combine the two half-reactions and cancel out the common species:

2ClO3^- + 16H^+ + 3Cl^- → 3Cl2 + 8H2O

e. In the acidic solution, the balanced redox equation Mn^2+ + BiO3^- → Bi^3+ + MnO4^- can be balanced as follows:

Reduction: BiO3^- → Bi^3+

Oxidation: Mn^2+ → MnO4^-

To balance the reduction half-reaction, add six H^+ ions to the left side:

BiO3^- + 6H^+ → Bi^3+ + 3H2O + 6e^-

To balance the oxidation half-reaction, add eight H^+ ions to the right side:

Mn^2+ → MnO4^- + 8H^+ + 5e^-

Multiply the reduction half-reaction by five and the oxidation half-reaction by two to equalize the electrons:

5(BiO3^- + 6H^+ → Bi^3+ + 3H2O + 6e^-)

2(Mn^2+ → MnO4^- + 8H^+ + 5e^-)

Combine the two half-reactions and cancel out the common species:

10BiO3^- + 60H^+ + 12Mn^2+ → 10Bi^3+ + 30H2O + 12MnO4^-

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Ask your parents or guardian for a tutor.

It's very dangerous to give someone your snap online that you don't know!

Have a nice day <3

Answer:

3fe s )+ 4h2o G )= fe3o4 s )+ 4h2 G

Explanation: 3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g) (i) Iron metal is getting oxidised. (ii) Water is getting reduced. (iii) Water is acting as reducing agent.

Explanation:

i make a search

Answer:

Molarity is moles per liter. You have one mole in 0.750 liters

Explanation:

There are approximately 0.689 moles of krypton in the gas tank.

To determine the number of moles of krypton in the gas tank, we can use the ideal gas law equation:

PV = nRT

Where:

P = pressure (in bar)

V = volume (in liters)

n = number of moles

R = ideal gas constant (0.0831 L·bar/(mol·K))

T = temperature (in Kelvin)

Given:

Pressure (P) = 5.80 bar

Volume (V) = 3.25 liters

Temperature (T) = 25.5 °C = 25.5 + 273.15 = 298.65 K

Using the ideal gas law equation, we can solve for the number of moles (n):

n = PV / RT

n = (5.80 bar * 3.25 L) / (0.0831 L·bar/(mol·K) * 298.65 K)

n ≈ 0.689 moles

Therefore, there are approximately 0.689 moles of krypton in the gas tank.

If the gas in the tank were oxygen instead of krypton, the answer would change because the molar mass of oxygen is different from that of krypton. The ideal gas law equation remains the same, but the value of n (number of moles) would be different since it depends on the molar mass of the gas. Oxygen has a molar mass of approximately 32 g/mol, while krypton has a molar mass of approximately 84 g/mol. So, the number of moles of oxygen in the gas tank would be different and can be calculated using the same ideal gas law equation, but substituting the molar mass of oxygen instead of krypton.

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Answer:

The population of the long-legged birds decreases.

Explanation:

The population of the short-legged birds increases whereas the population of  long-legged birds decreases due to availability of food in that environment. The long-legged birds feed on fish whose population decreases due to drought conditions so the population of long-legged birds also decreases while on the other hand, the population of short-legged birds increases or remain the same because they feed on the insects and the insects are available in large amount and less affected by the drought conditions.

The MW of rGFP in the band can be determined by comparing the mobility of the band with that of protein standards run on the same gel.

The Western blotting technique is utilized to identify and detect specific proteins in a sample of tissue or extract. If there is a band in the W2 lane of the Western blot, one can conclude that rGFP is present in that band. The physical protein structure of rGFP could not be inferred from the presence of a band in the W2 lane of the Western blot.  This requires additional analysis such as X-ray crystallography, nuclear magnetic resonance, or cryo-electron microscopy to analyze protein structure. MW is the molecular weight which can be determined using a molecular weight marker that runs in a parallel lane to the protein extract on the gel. In the Western blotting method, SDS-PAGE is typically used to separate proteins based on their size. The SDS-PAGE gel is calibrated with protein markers that have a known molecular weight. Hence, the MW of rGFP in the band can be determined by comparing the mobility of the band with that of protein standards run on the same gel.

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Answer:

CH3CH2CH3

(CH3)3CLi

CH3SNa

Explanation:

There are 5.50 equivalents of Mg^2+ present in a solution containing 2.75 mol of Mg^2+.

The concept of equivalents is used to quantify the number of reactive entities or charges present in a solution. In the case of Mg^2+, each Mg^2+ ion carries two positive charges, so it is necessary to determine the number of moles of Mg^2+ and then convert it to equivalents.

Given:

Number of moles of Mg^2+ = 2.75 mol

To calculate the equivalents, we use the relationship that one mole of Mg^2+ is equal to 2 equivalents of Mg^2+ (since each Mg^2+ ion carries two positive charges):

Equivalents of Mg^2+ = Number of moles of Mg^2+ * 2

Equivalents of Mg^2+ = 2.75 mol * 2

Equivalents of Mg^2+ = 5.50 equivalents

Therefore, in a solution containing 2.75 mol of Mg^2+, there are 5.50 equivalents of Mg^2+ present.

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Answer:

Chromatography

Distillation

Evaporation

Filtration

Explanation:

Chromatography involves solvent separation on a solid medium.

Distillation takes advantage of differences in boiling points.

Evaporation removes a liquid from a solution to leave a solid material.

Filtration separates solids of different sizes.

Mixtures can be physically separated by using methods that use differences in physical properties to separate the components of the mixture, such as evaporation, distillation, filtration and chromatography.

Answer:

i dont know i do this for points

Explanation:

Answer: The percentage by mass of sulphur in [tex]Al_2(SO_4)_3[/tex] is 9.36%

Explanation:

Mass percent of an element is the ratio of mass of that element by the total mass expressed in terms of percentage.

[tex]{\text {Mass percentage}}=\frac{\text {mass of sulphur}}{\text {Total mass}}\times 100\%[/tex]

Given: mass of sulphur = 32 g/mol

mass of [tex]Al_2(SO_4)_3[/tex] = 342 g/mol

Putting in the values we get:

[tex]{\text {Mass percentage}}=\frac{32g/mol}{342g/mol}\times 100\%=9.36\%[/tex]

The percentage by mass of sulphur in [tex]Al_2(SO_4)_3[/tex] is 9.36%

Answer:Divergent Boundaries(or Boundary)

Explanation:took the test and got it right

Answer:

V = 5000 mL

Explanation:

¡Hola!

En este caso, dado que la molaridad de una solution se calcula al dividir las moles por el volume de solución en litros, es posible calcular el volumen cuando se dividen las moles por la molaridad, tal y como se muestra a continuación:

[tex]M=\frac{n}{V}\\\\V=\frac{n}{M}[/tex]

Así, podemos reemplazar la molaridad y moles dadas para obtener:

[tex]V=\frac{1.5mol}{0.3mol/L}=5L[/tex]

Que en mililitros sería:

[tex]V=5L*\frac{1000mL}{1L}\\\\V=5000mL[/tex]

¡Saludos!

Answer:

[tex]C=0.91\frac{J}{g\°C}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it is possible to recall the equation to calculate the heat, Q, in these calorimetry problems as shown below:

[tex]Q=mC(T_f-T_i)[/tex]

Thus, given the absorbed heat, mass and temperatures, we can easily calculate the specific heat of the metal as shown below:

[tex]C=\frac{Q}{m(T_f-T_i)}[/tex]

Then, by plugging in we obtain:

[tex]C=\frac{1350J}{55.0g(47.0\°C-20.0\°C)} \\\\C=0.91\frac{J}{g\°C}[/tex]

Best regards!

Answer:

The amplitude of the speedboat waves were larger then the Breeze waves. The frequency of the speedboats waves were lower than the breeze waves. the speedboat waves had more of an effect on the boat, which tells us the speedboat waves had more energy.

Explanation:

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Answer:

speedboat waves artificail and waves by breezes natural

Explanation:

To determine the empirical formula of a compound, the given masses of the elements (carbon and oxygen) are used to calculate the moles of each element. The mole ratio between the elements is then determined to find the simplest whole number ratio, which represents the empirical formula.

First, we calculate the moles of carbon and oxygen using their respective molar masses. The molar mass of carbon (C) is approximately 12.01 g/mol, and the molar mass of oxygen (O) is approximately 16.00 g/mol.

Moles of carbon = 20.4 g / 12.01 g/mol ≈ 1.70 mol

Moles of oxygen = 54.4 g / 16.00 g/mol ≈ 3.40 mol

Next, we determine the simplest whole number ratio by dividing the number of moles of each element by the smallest number of moles. In this case, the smallest number of moles is approximately 1.70 mol (carbon).

Carbon: 1.70 mol / 1.70 mol ≈ 1

Oxygen: 3.40 mol / 1.70 mol ≈ 2

Therefore, the empirical formula for this compound is C1O2, which can be simplified to CO2.

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The platinum-rhodium catalyst used in most modern catalytic converters in automobiles is primarily employed for the oxidation of harmful pollutants. It facilitates the conversion of carbon monoxide (CO) and unburned hydrocarbons (HC) into carbon dioxide (CO2) and water (H2O).

The platinum-rhodium catalyst in catalytic converters is specifically designed to promote the oxidation reactions of carbon monoxide (CO) and unburned hydrocarbons (HC). These reactions are crucial for reducing the emission of harmful pollutants from automobile exhaust gases.

Carbon monoxide (CO) is a toxic gas produced during incomplete combustion. The platinum-rhodium catalyst assists in the oxidation of CO, converting it into carbon dioxide (CO2). This reaction is represented by the equation:

2 CO + O2 → 2 CO2

Unburned hydrocarbons (HC) are volatile organic compounds (VOCs) that contribute to smog formation. The platinum-rhodium catalyst aids in their oxidation, transforming them into carbon dioxide (CO2) and water (H2O). The general reaction can be expressed as:

HC + O2 → CO2 + H2O

The platinum-rhodium catalyst is essential in facilitating these oxidation reactions, promoting more complete combustion of harmful pollutants and reducing their negative environmental impact.

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Answer: The mass

Explanation: ability to rust, flammability, and ability to combust are chemical properties.

A low vapor pressure indicates the presence of weak intermolecular forces in a liquid. The correct answer is: c.

The strength of intermolecular forces in a liquid determines the vapor pressure of the liquid. A liquid with strong intermolecular forces will have a low vapor pressure, while a liquid with weak intermolecular forces will have a high vapor pressure.

This is because the molecules in a liquid with weak intermolecular forces are more likely to escape from the surface of the liquid and enter the gas phase.

The other options are incorrect because they are all properties that indicate the presence of strong intermolecular forces in a liquid. A high boiling point indicates that a large amount of energy is required to overcome the intermolecular forces and vaporize the liquid.

A high surface tension indicates that the molecules in the liquid are strongly attracted to each other and to the surface of the liquid. A low heat of vaporization indicates that a small amount of energy is required to overcome the intermolecular forces and vaporize the liquid.

Therefore, the only property that indicates the presence of weak intermolecular forces in a liquid is a low vapor pressure.

Therefore, the correct option is C, a low vapor pressure.

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it should be carbon dioxide and water

Answer:

it is closer to the south pole

Explanation:

Answer:

a beam of electrons emitted from the cathode of a high-vacuum tube.

Explanation:

Cathode rays (also known as electron beam or e-beam) are electron waves that can be used in vacuum tubes.

Answer:

10.58 J/g-°C

Explanation:

To find the specific heat capacity of the metal, you need to know how much heat was lost when it reacted with water.

You know that there are 55.4 g of water, the initial temp. of water is 25°C, and the final temp. (the mixture's temp.) is 63.4°C.

You should also know that the specific heat capacity of water is 4.186 J/g-°C.

Plug this into the equation the q=mcΔT.

q = (55.4 g)(4.186 J/g-°C)(63.4°C - 25°C)

q = 8905.12896 J

If 8905.12896 J was gained by the water, then 8905.12896 J must have been lost from the metal.

You know that there are 23 g of the metal and that its initial temp. is 100°C.

Plug this information into q=mcΔT.

8905.12896 J = (23 g)(C)(63.4°C - 100°C)

C = 10.58 J/g-°C

*When you plug all of this into the calculator, it will result in a negative number but keep in mind that heat was LOST by the metal so 8905.12896 J  is essentially negative. So the negative cancels out.*

Answer:

The final balanced equation is : 2C2H2+5O2→4CO2+2H2O.

Answer:

May be the instrument is incorrect or may be error in it.

Explanation:

The copper have not been detected by this test because the test may be not for the detection of copper, may be it is used for identification of another minerals. If there is copper in the lake sample but can't be detected in the test so it means that the instrument which is used for detection is not the right one  or having error in that instrument. Every mineral has a specific type of instrument that detect its presence, if we use incorrect instrument for the mineral then we can't detect the presence of that specific mineral.

Answer:

5 mL

Explanation:

As this is a problem regarding dilutions, we can solve it using the following formula:

Where subscript 1 refers to the initial concentration and volume, while 2 refers to the final C and V. Meaning that in this case:

We input the data:

And solve for V₁:

Answer:

5 mL of 5.0 M H₂SO₄ (aq) are needed to prepare 100 mL of 0.25 M H₂SO₄ (aq).

Explanation:

In chemistry, dilution is the reduction of the concentration of a chemical in a solution.

Then, dilution consists of preparing a less concentrated solution from a more concentrated one, and it consists simply by adding more solvent to the same amount of solute. That is, the amount or mass of the solute is not changed, but the volume of the solvent varies: as more solvent is added, the concentration of the solute decreases, since the volume (and weight) of the solution increases.

A dilution is calculated by the expression:

Ci*Vi = Cf*Vf

where:

In this case, you know:

Replacing:

5 M*Vi = 0.25 M* 100 mL

Solving:

[tex]Vi= \frac{0.25 M*100 mL}{5 M}[/tex]

Vi= 5 mL

5 mL of 5.0 M H₂SO₄ (aq) are needed to prepare 100 mL of 0.25 M H₂SO₄ (aq).