Which is the next logical step in balancing the given equation?
CS2(l) + Cl2(g) CCl4(l) + S2Cl2(l)
A.) Place the coefficient 2 in front of sulfur dichloride (S2Cl2).
B.)Place the coefficient 3 in front of the chlorine molecule.
C.)Leave the equation alone as it is already balanced.
D.) Place the coefficient 4 in front of carbon disulfide(CS2).
E.)Place the coefficient 2 in front of carbon tetrachloride(CCl4).

Answer: The correct answer is [tex]3.6\times 10^{-6}[/tex]

Explanation:

We are given:

Acid dissociation constant [tex](k_a)=2.8\times 10^{-9}[/tex]

Water dissociation constant [tex](k_w)=1\times 10^{-14}[/tex]

To calculate the base dissociation constant for the conjugate base, we use the equation:

[tex]k_w=k_a\times k_b[/tex]

where,

[tex]k_b[/tex] = base dissociation constant

Putting values in above equation, we get:

[tex]10^{-14}=2.8\times 10^-9}\times k_b\\\\k_b=3.6\times 10^{-6}[/tex]

Hence, the correct answer is [tex]3.6\times 10^{-6}[/tex]

Explanation:

Molality is defined as the number of moles of solute per kg of solution. That is, molality depends on the mass.  

Whereas molarity is defined as the number of moles per liter of solution. That is, molarity depends on the volume of solution.

Also, when there is change in temperature then volume of solution changes but mass does not change.

Thus we can conclude that investigating changes in colligative properties involves the measurement of a solution's concentration in molality instead of molarity because mass does not change whereas volume does change.

HCHO2 and NaCHO2, and CH3NH2 and CH3NH3Cl can make buffer solutions. HCl and NaCl, and NH3 and NaOH cannot make buffer solutions.

1. HCHO2 is formic acid, a weak acid, while NaCHO2 is the salt made from the anion of the weak acid (the formate ion [CHO2]). The combination of these two solutes would make a buffer solution.

2. HCl is a strong acid, not a weak acid, so the combination of these two solutes would not make a buffer solution.

3. CH3NH2 is methylamine, which is like NH3 with one of its H atoms substituted with a CH3 group. Because it is not listed in Table 12.8.1, we can assume that it is a weak base. The compound CH3NH3Cl is a salt made from that weak base, so the combination of these two solutes would make a buffer solution.

4. NH3 is a weak base, but NaOH is a strong base. The combination of these two solutes would not make a buffer solution.

This is a problem based on concept of first law of thermodynamics

if a body is giving heat it will be absorbed by some other body or surrounding.

Now we have taken metal at low temperature and water at high temperature

So water will lose some energy and the same amount of energy will be gained by metal. The exchange will continue untill both reach the same temperature.

So answer is

The energy gained by the metal equals the energy lost by the water.

Answer: Option (D) is the correct answer.

Explanation:

When there is minimum use of energy and we are still able to preserve resources then it means we are cutting the demand for energy.

Therefore, when we are treating water with chlorine then it means without any electricity or energy we are cleaning the water. But at the same time chlorinated water is harmful to use unless chlorine is dissolved in limiting quantity.

Whereas energy efficient appliances will use the electricity efficiently.

Landfill will require the use of machines or man power therefore, energy will be used.

On the other hand, when rules are framed against the use of chlorofluorocarbons then there will be lesser depletion of ozone layer. Hence, use of energy will be minimum.

Thus, we can conclude that legislation against the use of chlorofluorocarbons has helped preserve resources while cutting the demand for energy.

Answer: Oxygen is the second most abundant gas in our atmosphere. ap

Explanation:

The balanced nuclear equation for the beta decay of carbon-11 is ⁶₁₁C → ⁶₁₂N + ⁰⁻₁e, which indicates carbon-11 transforming into nitrogen-11 and emitting a beta particle (electron).

The beta decay of carbon-11 is a type of radioactive decay where a beta particle (an electron) is emitted from an atomic nucleus. Writing a balanced nuclear equation for this process involves showing the original nucleus, the emitted beta particle, and the resulting daughter nucleus. In beta decay, a neutron in the nucleus is transformed into a proton and an electron. The electron is ejected from the nucleus as the beta particle, and the atomic number of the nucleus increases by one while the mass number remains the same.

The balanced nuclear equation for the beta decay of carbon-11 is:

⁶₁₁C → ⁶₁₂N + ⁰⁻₁e

Here, carbon-11 (⁶₁₁C), with 6 protons and 5 neutrons, decays to nitrogen-11 (⁶₁₂N), which has 7 protons and 4 neutrons, by emitting a beta particle (⁰⁻₁e, also represented as β-). This process increases the atomic number by one, from carbon (6) to nitrogen (7), while the mass number remains 11.

Final answer:

The balanced nuclear equation for the beta decay of carbon-11 is 11C → 11B + β + γ. The daughter isotope in this decay is boron-11 (11B).

Explanation:

The balanced nuclear equation for the beta decay of carbon-11 is:

11C → 11B + β + γ

The daughter isotope in this decay is boron-11 (11B).

When aqueous solutions of [tex]\rm NiBr_2\;and\;AgNO_3[/tex]  are mixed, a precipitate form.

A precipitation can occur when two solutions containing different salts are mixed, and a cation/anion pair in the resulting combined solution forms an insoluble salt; this salt then precipitates out of solution.

In the given combination,

[tex]\rm K_2SO_4\;and\;CrCl_3[/tex] : The reaction will result in Group 1A compounds which are soluble.

[tex]\rm NaI\;and\;KBr[/tex] : The reaction will result in Group 1A compounds which are soluble.

[tex]\rm Li_2CO_3\;and\;CSi[/tex] : carbonates with group 1A elements becomes soluble.

[tex]\rm NiBr_2\;and\;AgNO_3[/tex] : The product will be a halide with Ag which are insoluble in water, and thus precipitates out.

When aqueous solutions of [tex]\rm NiBr_2\;and\;AgNO_3[/tex]  are mixed, a precipitate form.

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To find the density of a liquid with a mass of 22.00 grams and a volume of 10.0 ml, divide the mass by the volume to get a density of 2.20 g/ml.

To calculate the density of a liquid, you can use the formula:

Density = mass / volume

In this case, the mass of the liquid is 22.00 grams and the volume is 10.0 ml (which can also be written as 10.0 cm³ as 1 ml is equivalent to 1 cm³). Plugging these values into the formula gives us:

Density = 22.00 g / 10.0 ml = 2.20 g/ml

Therefore, the density of the liquid is 2.20 g/ml.

Answer:  Carbonyl is the functional group.

Explanation:  Aldehyde is a functional group containing the group -CHO .

Amide is the functional group which contains the group -CONH2.

Carboxyl is the functional group which contains the group -COOH.

And Hydroxyl is the functional group which contains the group -OH.

The freezing point of the solution is 0.88°C

The melting point of a substance which is also the same as the freezing point is the temperature at which it changes from a solid state to a liquid state at a specific pressure. At the melting point, the substance absorbs heat energy, increasing its temperature and causing the solid structure to break down

We have that;

Number of moles of naphthalene = 10 g/128 g/mol

= 0.078 moles

Mass of benzene = Density * volume

= 0.877 g/cm3 * 100.0 mL

= 87.7 g or 0.0877 Kg

Where;

ΔT = K m i

ΔT = 5.12 *  0.078 moles/ 0.0877 Kg * 1

= 4.6°C

ΔT = Freezing point of pure solvent - Freezing point of solution

Freezing point of solution = 5.48°C - 4.6°C

= 0.88°C

Missing parts

Calculate the freezing point and melting point of a solution containing 10.0 g of naphthalene (C10H8) in 100.0 mL of benzene. Benzene has a density of 0.877 g/cm3

Answer:

c

Explanation:

The balanced thermochemical equation for the decomposition of methanol (CH3OH(g)) to form CH4(g) and O2(g) is CH3OH(g) -> CH4(g) + O2(g) + 252.8 kJ/mol of O2 formed.

The balanced thermochemical equation for the decomposition of methanol (CH3OH(g)) to form CH4(g) and O2(g) can be written as:

CH3OH(g) -> CH4(g) + O2(g) + 252.8 kJ/mol of O2 formed.

To convert 2,150 mmHg to kPa, divide by the conversion factor 7.5 mmHg, resulting in approximately 286 kPa.

A conversion factor is a mathematical ratio that allows you to convert from one unit of measurement to another. It is often used to change units within the same system (e.g., converting inches to centimeters) or between different systems (e.g., converting miles to kilometers) by multiplying or dividing by the appropriate factor.

To convert from millimetres of mercury (mmHg) to kilopascals (kPa), you can use the conversion factor:

1 kPa = 7.5 mmHg

First, divide 2,150 mmHg by the conversion factor:

= 2,150 mmHg / 7.5 mmHg

= 286.67 kPa

Rounding to the nearest hundredth, the answer is approximately 286 kPa.

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Answer:

[tex]\rm PbSO_4[/tex] will precipitate in the given reaction.

Explanation:

The reaction takes place between lead nitrate [tex]\rm Pb(NO_3)_2[/tex] and potassium sulphate [tex]\rm K_2SO_4[/tex].

Now, the reaction will be a double displacement reaction in which the ions will exchange between lead nitrate [tex]\rm Pb(NO_3)_2[/tex] and potassium sulphate [tex]\rm K_2SO_4[/tex].

So, the chemical reaction can be written as,

[tex]\rm {Pb(NO_3)_2+K_2SO_4}\rightarrow PbSO_4+2KNO_3[/tex]

In the above equation, [tex]\rm PbSO_4[/tex] is a precipitate which will be settled down in the  bottom and [tex]\rm KNO_3[/tex] will form an aqueous solution.

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The mass of H2(g) collected is approximately calculated as 0.582 grams. This result is found by subtracting the water's vapor pressure from the total pressure, applying the ideal gas law to find the number of moles, and converting moles to grams.

To calculate the mass of H2(g) collected, we'll first need to calculate the pressure of the dry hydrogen gas. The pressure of a gas collected over water is a sum of the pressure of the dry gas and the vapor pressure of the water. Hence, the pressure of dry H2(g) is the total pressure minus the water vapor pressure, so 685 mmHg - 18.6 mmHg = 666.4 mmHg.

Next, we convert this pressure into atmospheres: 666.4 mmHg * (1 atm / 760 mmHg) = 0.877 atm (approximately). We'll also need to convert the temperature to Kelvin: 21ºC + 273 = 294 K.

Then, we can use the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L*atm/K*mol), and T is temperature. Solving for n, n = PV / RT, we substitute the known values to get n = (0.877 atm * 7.80 L) / (0.0821 L*atm/K*mol * 294 K) = 0.291 mol.

Last, we turn moles of H2 into grams. As the molar mass of H2(g) is approximately 2 g/mol, the mass of H2(g) collected is 0.291 mol * 2 g/mol = 0.582 grams.

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The correct answers are:

1. B. Mg loses two electrons.

When Mg and Br combine, 2 atoms of Br attaches itself to Mg. The chemical reaction is:

Mg + Br ---> MgBr2

Since Br is more electronegative than Mg, then Mg loses an electron per Br therefore losing 2 electrons.

2. D. An atom that gains electrons must be attracted to an atom that loses electrons.

An ionic bond is formed when one molecule is more electronegative than the other molecule which results in gaining and losing of electrons. The more electronegative molecule gains electron while the less electronegative loses electron.

In the formation of an ionic bond between magnesium and bromine, magnesium loses two electrons to become a Mg₂ + ion, and these oppositely charged ions attract each other to form the bond.

Understanding Ionic Bonds between Magnesium and Bromine:

1. When an ionic bond forms between magnesium (Mg) and bromine (Br), the process involves a transfer of electrons. Specifically, Mg loses two electrons (Option B) to achieve a stable electron configuration similar to the nearest noble gas. As a result, magnesium becomes a Mg₂+ ion.

2. For an ionic bond to form, there must be a transfer of electrons from one atom to another, leading to the formation of ions with opposite charges, which are then attracted to each other (Option D). An atom with a lower electronegativity, like magnesium, will lose electrons and become a positively charged cation. Conversely, an atom with higher electronegativity will gain electrons and become a negatively charged anion, such as bromine becoming Br-.