In a science demonstration, a teacher mixed zinc (Zn) with hydrogen chloride (HCl) in a flask and quickly attached a balloon over the mouth of the flask. Bubbles formed in the solution and the balloon inflated.
What most likely occurred during this demonstration?

a.The Zn and HCl both retained their identity.
b.Either Zn or HCl, but not both, retained its identity.
c.Evaporation of one of the substances occurred.
d.One or more new substances formed.

Answers

Answer 1

Answer:

a. The Zn and HCl both retained their identity.


Related Questions

Re-read the Topic 2 Learning Activities titled “Glycolysis” and “Overview of Photosynthesis”. What makes these necessary fundamental processes? Use an argument from the reading to support your answer. In what ways are these two processes similar? How are they different?

Answers

Glycolysis and photosynthesis are fundamental processes that are necessary for the survival of living organisms. They are similar in that they both involve the conversion of energy, but differ in the source of energy used, the location of the process, and the requirement for oxygen.

Glycolysis and photosynthesis are two necessary fundamental processes. Glycolysis is a metabolic pathway that occurs in the cytoplasm of cells. The glycolysis process is necessary because it produces ATP, which is the energy required for all cellular activities.

The energy is produced by breaking down glucose into two pyruvate molecules.Photosynthesis is the process by which green plants make their food. During photosynthesis, light energy is converted into chemical energy, which is stored in glucose molecules. This process is also necessary as it provides food and oxygen for most living organisms to survive.In terms of similarities, both glycolysis and photosynthesis are processes that involve the conversion of energy.

In glycolysis, glucose is converted into pyruvate and ATP, while in photosynthesis, light energy is converted into chemical energy. Both processes are also vital to the survival of living organisms.The primary difference between the two processes is the source of energy used. Glycolysis uses glucose as the primary energy source while photosynthesis uses light energy from the sun.

Glycolysis occurs in the cytoplasm of cells while photosynthesis takes place in the chloroplasts of plant cells. Glycolysis is an anaerobic process that does not require oxygen, while photosynthesis is an aerobic process that requires oxygen and releases it as a byproduct.

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Mention three significant of water in coal fired power station

Answers

Water in coal-fired power stations is used for cooling, steam generation, and pollution control, including capturing sulfur dioxide and cooling exhaust gases. Efficient water recycling helps minimize environmental impact.

Water plays a critical role in coal-fired power stations. The power stations need large quantities of water for a variety of purposes. Water is primarily used to cool the power plant, maintain a safe temperature in the boilers, and also to generate steam. In this context, this answer will discuss three significant uses of water in coal-fired power stations. Significant uses of water in coal-fired power stations1. Cooling: Power stations require water for cooling purposes. When water is used for cooling, it absorbs the heat produced by the combustion process. Cooling towers are responsible for releasing the heated water, which is then reused.2. Steam generation: Water is required to generate steam, which is used to rotate turbines and generate electricity. The water used to generate steam must be treated to prevent the accumulation of harmful minerals, which can damage the power plant.3. Pollution control: Water is utilized to reduce air pollution. Flue gas desulfurization systems utilize water to capture sulfur dioxide from power plants. Water is also used to cool exhaust gases that are produced during combustion.In conclusion, the three significant uses of water in coal-fired power stations include cooling, steam generation, and pollution control. These processes require large amounts of water, which is why coal-fired power stations are often located near water sources. By recycling water, power stations can conserve water and minimize their environmental impact.

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Calculate the Standard Free Energy Change at 25℃ given the Equilibrium constant of 1.3 × 104.

Answers

The standard free energy change at 25℃ is -2.48 × 10⁴ J/mol.

The equation linking Gibbs free energy change and equilibrium constant is given by the following equation:

ΔG° = -RT ln K(where, ΔG° is the standard free energy change, R is the gas constant, T is the temperature in Kelvin and K is the equilibrium constant)

Substituting the given values:Equilibrium constant, K = 1.3 × 10⁴

Standard temperature, T = 25℃ = 298K

Substituting the values in the equation of Gibbs free energy change:

ΔG° = -RT ln K=-8.31 J K⁻¹ mol⁻¹ × 298 K × ln 1.3 × 10⁴

      = -8.31 J K⁻¹ mol⁻¹ × 298 K × 9.480

     = -2.48 × 10⁴ J/mol (Approx)

Therefore, the standard free energy change at 25℃ is -2.48 × 10⁴ J/mol.

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100 POINTS!!!

What is the average rate of the reaction over the entire course of the reaction?
1.6 × 10−3 (?)
1.9 × 10−3 (?)
2.0 × 10−3 (X)
2.2 × 10−3 (X)

Answers

Answer:

b.  1.9 × 10-3

Explanation:

Answer:1.9x10-3

Explanation:

average

The composition of a compound is 28.73% K, 1.48% H, 22.76% P, and 47.03% O. The molar mass of the
compound is 136.1 g/mol.
I

Answers

The compound has an empirical formula of [tex]K_2H_2P_2O_8[/tex] and a molecular formula of [tex]K_2HPO_4[/tex].

The given compound has a percent composition of K = 28.73%, H = 1.48%, P = 22.76%, and O = 47.03%. Its molar mass is 136.1 g/mol. To determine its molecular formula, we need to find its empirical formula and calculate its molecular formula from its empirical formula.The empirical formula is the smallest whole number ratio of atoms in a compound. It can be determined by converting the percent composition of the elements into their respective moles and dividing each by the smallest number of moles calculated. The moles of K, H, P, and O in 100 g of the compound are: K = 28.73 g x (1 mol/39.1 g) = 0.734 molH = 1.48 g x (1 mol/1.01 g) = 1.46 molP = 22.76 g x (1 mol/30.97 g) = 0.736 molO = 47.03 g x (1 mol/16.00 g) = 2.94 molDividing each by the smallest number of moles gives the following ratios: K = 0.734/0.734 = 1H = 1.46/0.734 = 2P = 0.736/0.734 = 1.002O = 2.94/0.734 = 4. The empirical formula of the compound is [tex]K_2H_2P_2O_8[/tex]. To calculate the molecular formula, we need to determine the factor by which the empirical formula should be multiplied to obtain the molecular formula. This can be done by comparing the molar mass of the empirical formula to the molar mass of the compound.The molar mass of [tex]K_2H_2P_2O_8[/tex] is: [tex]M(K_2H_2P_2O_8)[/tex] = (2 x 39.1 g/mol) + (2 x 1.01 g/mol) + (2 x 30.97 g/mol) + (8 x 16.00 g/mol) = 276.2 g/mol. The factor by which the empirical formula should be multiplied is: M(molecular formula)/M(empirical formula) = 136.1 g/mol/276.2 g/mol = 0.4935. The molecular formula is obtained by multiplying the empirical formula by this factor: [tex]K_2H_2P_2O_8 * 0.4935 = K_2HPO_4[/tex]. Therefore, the molecular formula of the compound is [tex]K_2HPO_4[/tex].The molecular formula of the given compound having a composition of 28.73% K, 1.48% H, 22.76% P, and 47.03% O with a molar mass of 136.1 g/mol is [tex]K_2HPO_4[/tex]. The empirical formula of the compound is [tex]K_2H_2P_2O_8[/tex]. The compound's molecular formula is calculated by determining the factor by which the empirical formula should be multiplied to obtain the molecular formula. The factor is M(molecular formula)/M(empirical formula) = 136.1 g/mol/276.2 g/mol = 0.4935. The molecular formula of the compound is obtained by multiplying the empirical formula by this factor, resulting in the molecular formula [tex]K_2HPO_4[/tex].

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The correct question would be as

The composition of a compound is 28.73% K. 1.48% H, 22.76% P, and 47.03% O. The molar mass of the compound is 136.1 g/mol. What is the Molecular Formula of the compound?

[tex]KH_2PO_4\\KH_3PO_4\\K_2H_4P_20_{12}\\K_2H_3PO_6[/tex]

Calculate the mass of wire that reacted to silver nitrate solution Mass being 1.52 of copper before reaction

Answers

The mass of wire that reacted to silver nitrate solution is 5.15 grams.

To calculate the mass of the wire that reacted with silver nitrate solution, we need to consider the stoichiometry of the reaction. The reaction between copper and silver nitrate can be represented by the following equation:

Cu + 2AgNO3 → Cu(NO3)2 + 2Ag

According to the equation, one mole of copper reacts with two moles of silver nitrate to form one mole of copper(II) nitrate and two moles of silver.

Given that the mass of copper before the reaction is 1.52 grams, we can calculate the molar mass of copper using its atomic mass, which is 63.55 grams/mol.

1.52 g of copper is equal to 1.52 g / 63.55 g/mol = 0.0239 moles of copper.

Since the reaction stoichiometry is 1:2 between copper and silver, the moles of copper reacting would be equal to half of the moles of silver formed.

Therefore, the moles of silver formed would be 0.0239 moles x 2 = 0.0478 moles.

To find the mass of silver, we multiply the moles of silver by its molar mass, which is 107.87 grams/mol:

Mass of silver = 0.0478 moles x 107.87 g/mol = 5.15 grams.

Hence, the mass of wire that reacted to silver nitrate solution is 5.15 grams.

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I think it is the question:

A copper wire with a mass of 1.52 grams reacted with silver nitrate solution. If the balanced chemical equation and the molar ratio between copper and silver nitrate are provided, how can you determine the mass of wire that reacted?

Which element is the mostvreactive, based on the data?

A. Element J
B. Element K
C. Element L
D. Element I​

Answers

The most reactive element based on the given data among the given options is option c) Element J.  

This can be determined based on their placement on the periodic table. The reactivity of an element is dependent on its position on the periodic table, particularly its electron configuration and the number of valence electrons it has. For instance, elements located in the top left corner of the periodic table are typically the most reactive.

They have fewer electrons in their outermost shell and have a tendency to lose them or combine with other elements in order to obtain a full outer shell or achieve stability.In this case, Element J is most likely located in the far left of the periodic table, most likely in the alkali metals group, which contains some of the most reactive metals.

Alkali metals are highly reactive because they only have one valence electron, making it easy for them to give it up and form positive ions. As a result, Element J is the most reactive among the given elements.The correct answer is c.

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Starting with 0.3500 mol CO(g) and 0.05500 mol COCl2(g) in a 3.050 L flask at 668 K, how many moles of CI2(g) will be present at equilibrium? CO(g) + Cl2(8)》COCl2(g)
Kc= 1.2 x 10^3 at 668 K

Answers

At equilibrium, the number of moles of Cl2(g) present is approximately 347.37 mol.

To determine the number of moles of Cl2(g) at equilibrium, we need to use the given equilibrium constant (Kc) and set up an ICE table to track the changes in the reactants and products.

The balanced equation for the reaction is:

CO(g) + Cl2(g) ⇌ COCl2(g)

Let's set up the ICE table:

  CO(g)     +     Cl2(g)     ⇌     COCl2(g)

Initial: 0.3500 0.05500 0

Change: -x -x +x

Equilibrium: 0.3500 - x 0.05500 - x x

Using the equilibrium concentrations in the ICE table, we can write the expression for the equilibrium constant (Kc) as:

Kc = [COCl2(g)] / [CO(g)][Cl2(g)]

Substituting the values into the equation, we have:

1.2 × 10^3 = (0.05500 - x) / [(0.3500 - x)(0.05500 - x)]

Simplifying the equation, we can cross-multiply and rearrange:

1.2 × 10^3 × (0.3500 - x)(0.05500 - x) = 0.05500 - x

Expanding and rearranging, we get:

0 = (1.2 × 10^3 × 0.05500 - 1.2 × 10^3x + 0.05500x) - x

Simplifying further:

0 = 66 - 1.245x + 0.05500x - x

0 = 66 - 0.19x

0.19x = 66

x = 66 / 0.19

x ≈ 347.37

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A scientist wants to make advances in the way skin cancer patients are
treated. What is something she should do first?

Answers

To make advances in the way skin cancer patients are treated, a scientist should first conduct thorough research and gather relevant information.

Review existing literature: The scientist should study the current scientific literature on skin cancer treatment, including the latest research, clinical trials, and treatment options. This will provide a foundation of knowledge and help identify gaps or areas that need improvement.

Understand the current challenges: It is important for the scientist to have a comprehensive understanding of the challenges faced by skin cancer patients and healthcare providers in existing treatment methods. This can involve analyzing the limitations of current therapies, side effects, recurrence rates, and patient outcomes.

Identify unmet needs: Through research and engagement with dermatologists, oncologists, and patients, the scientist should identify specific unmet needs in skin cancer treatment. This can include areas such as early detection methods, personalized therapies, targeted drug delivery, or improving the effectiveness of existing treatments.

Collaborate with multidisciplinary teams: Skin cancer treatment often requires collaboration between various specialists, including dermatologists, oncologists, surgeons, and researchers. The scientist should establish collaborations with experts from different disciplines to gain diverse perspectives and insights.

Conduct preclinical and clinical research: Once the specific objectives are identified, the scientist should design and conduct preclinical studies to explore potential treatment approaches. This can involve laboratory experiments, animal models, and in vitro studies.

Evaluate safety and efficacy: During clinical trials, the scientist should rigorously evaluate the safety and efficacy of the new treatment approach. This includes monitoring patient responses, side effects, and overall outcomes.

Analyze and disseminate results: The scientist should carefully analyze the data collected during the research and communicate the findings through scientific publications, conferences, and collaborations.

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organic functional groups that are found in morphine but not in cannabinol

Answers

Organic functional groups found in morphine but not in cannabinol include:

1. Phenol group: Morphine contains a phenol group (-OH) attached to an aromatic ring, which is absent in cannabinol.

2. Ether group: Morphine possesses an ether functional group (-O-) in its structure, while cannabinol does not have this group.

3. Amino group: Morphine contains an amino group (-NH2), which is not present in cannabinol.

4. Tertiary amine group: Morphine has a tertiary amine group (-NR3), whereas cannabinol lacks this functional group.

5. Ester group: Ester functional groups (-COO-) are found in some morphine derivatives, but cannabinol does not possess this group.

It's important to note that while these functional groups differentiate morphine from cannabinol, the effects and properties of these compounds are determined by their overall chemical structure, not just the presence or absence of specific functional groups.
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