Many people have said that cold water boils faster than hot water. This is not true. In fact, it’s been said so many times that most people believe it to be a fact. Postulate a reason for why this may have been thought to be true. Is there any scientific evidence backing this claim at all? Please explain your reasoning.

The empirical formula of the compound is C₂H₁₆S₂N₃O.

The moles of each element is as follows::

For CO₂:

Carbon (C) has a molar mass of 12.01 g/mol.

Oxygen (O) has a molar mass of 16.00 g/mol.

Moles of C in CO₂ = 2.08 g / 12.01 g/mol = 0.173 moles

Moles of O in CO₂ = 2.08 g / 16.00 g/mol = 0.130 moles

For H₂O:

Hydrogen (H) has a molar mass of 1.01 g/mol.

Oxygen (O) has a molar mass of 16.00 g/mol.

Moles of H in H₂O = 1.28 g / 1.01 g/mol = 1.27 moles

Moles of O in H₂O = 1.28 g / 16.00 g/mol = 0.080 moles

For SO₃:

Sulfur (S) has a molar mass of 32.06 g/mol.

Oxygen (O) has a molar mass of 16.00 g/mol.

Moles of S in SO₃ = 4.77 g / 32.06 g/mol = 0.149 moles

Moles of O in SO₃ = 4.77 g / 16.00 g/mol = 0.298 moles

For HNO₃:

Hydrogen (H) has a molar mass of 1.01 g/mol.

Nitrogen (N) has a molar mass of 14.01 g/mol.

Oxygen (O) has a molar mass of 16.00 g/mol.

Moles of H in HNO₃ = 3.48 g / 1.01 g/mol = 3.45 moles

Moles of N in HNO₃ = 3.48 g / 14.01 g/mol = 0.248 moles

Moles of O in HNO₃ = 3.48 g / 16.00 g/mol = 0.217 moles

The simplest whole-number ratio of the elements will be:

Carbon: 0.173 moles / 0.080 moles ≈ 2.16

Hydrogen: 1.27 moles / 0.080 moles ≈ 15.88

Sulfur: 0.149 moles / 0.080 moles ≈ 1.86

Nitrogen: 0.248 moles / 0.080 moles ≈ 3.10

Oxygen: 0.080 moles / 0.080 moles = 1

Therefore, the empirical formula is C₂H₁₆S₂N₃O.

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The correct formula for the compound that forms between aluminum (Al) and sulfur (S) is (d) Al₂S₃.

The correct option is E, none of the above.

The reaction between aluminum (Al) and sulfur (S) is an example of a chemical reaction known as a combination or synthesis reaction. In this reaction, aluminum and sulfur combine to form a

.

The chemical equation for the reaction is:

2Al + 3S → Al₂S₃

Aluminum has a 3+ charge (Al³⁺) and sulfur has a 2- charge (S²⁻). In order to balance the charges, two aluminum ions (Al³⁺) are needed to combine with three sulfur ions (S²⁻), resulting in the formula Al₂S.

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The final broth temperature will be approximately 38.4°C.

When mixing two substances with different temperatures, we can use the principle of conservation of energy. The energy lost by the hot substance (broth) is equal to the energy gained by the cold substance (water), assuming no energy is lost to the surroundings. This can be expressed using the equation:

Q_lost = Q_gained

The energy lost by the broth can be calculated using the formula:

Q_lost = m_broth * c_broth * (T_final - T_initial)

where m_broth is the mass of the broth, c_broth is its specific heat capacity, T_final is the final temperature, and T_initial is the initial temperature of the broth.

Similarly, the energy gained by the water can be calculated using:

Q_gained = m_water * c_water * (T_final - T_initial)

Since the two substances reach thermal equilibrium, we can set Q_lost equal to Q_gained:

m_broth * c_broth * (T_final - T_initial) = m_water * c_water * (T_final - T_initial)

Plugging in the given values and solving for T_final, we find that the final temperature of the broth is approximately 38.4°C.

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In the balanced equation [tex]2C_{2} H_{6}[/tex] + [tex]7 O_{2}[/tex] --> [tex]4 CO_{2}[/tex] + [tex]6H_{2}O[/tex] if 21 g of [tex]C_{2} H_{6}[/tex] reacts with 32 g O₂, C₂H6 is the limiting reactant.

To determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometric ratio in the balanced equation.

Let's calculate the number of moles for each reactant using their molar masses:

For [tex]C_{2} H_{6}[/tex] (ethane):

Molar mass of [tex]C_{2} H_{6}[/tex] = 2(12.01 g/mol) + 6(1.01 g/mol) = 30.07 g/mol

Number of moles of C₂H6 = 21 g / 30.07 g/mol ≈ 0.698 mol

For O₂ (oxygen):

Molar mass of O₂ = 2(16.00 g/mol) = 32.00 g/mol

Number of moles of O₂ = 32 g / 32.00 g/mol = 1.00 mol

Next, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation:

2 moles of [tex]C_{2} H_{6}[/tex] react with 7 moles of O₂ to produce 4 moles of CO₂ and 6 moles of H₂O.

From the given amounts, we have:

0.698 mol [tex]C_{2} H_{6}[/tex] and 1.00 mol O₂.

Using the stoichiometric ratio, we can calculate the expected amount of CO₂ and H₂O produced for each reactant:

For C₂H6:

Expected moles of CO₂ = 0.698 mol C₂H6 * (4 mol CO₂ / 2 mol C₂H6) = 1.396 mol CO₂

For O₂:

Expected moles of CO₂ = 1.00 mol O₂ * (4 mol CO₂ / 7 mol O₂) ≈ 0.571 mol CO₂

Comparing the expected moles, we see that the calculated amount of CO₂ is greater when used [tex]C_{2} H_{6}[/tex] as the limiting reactant. Therefore, the limiting reactant in this reaction is [tex]C_{2} H_{6}[/tex].

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Ribosomes are attached to the membrane of the ER, making it “rough.” The RER is also attached to the nuclear envelope that surrounds the nucleus. This direct connection between the perinuclear space and the lumen of the ER allows for the movement of molecules through both membranes.Answer:

Explanation:

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