Please help - Alevel questions on chapter 6 kinetics a level chemistry

Answer:

\(c=5.14\ J/kg ^{\circ} C\)

Explanation:

Given that,

Heat absorbed by a sample, Q = 228 J

Mass of a sample, m = 706 g = 0.706 kg

Initial temperature is 26 °C and final temperature is 88.8°C

We need to find the heat absorbed by the sample. The heat absorbed by an object is given by :

\(Q=mc\Delta T\\\\\text{Where c is specific heat of sample}\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{228\ J}{0.706\ kg(88.8-26)^{\circ} C}\\\\c=5.14\ J/kg ^{\circ} C\)

So, the specific heat of the sample is \(5.14\ J/kg ^{\circ} C\).

Answer:

c) 12

Explanation:

electron = 23-11 = 12

Answer:

D- 11

Explanation:

I know cause thats what the test said. Not sure how to do this stuff we will never use thats why im on brainly.

The first person to use heating processes to preserve foods was Nicholas Appert. He discovered that food can be kept from spoiling by heating it and sealing it in airtight containers.

Nicholas Appert, a French confectioner, invented the method of canning food in 1809. He conducted a series of experiments to find a way to preserve food for long periods without it spoiling. Appert discovered that food can be kept from spoiling by heating it and sealing it in airtight containers.

He used glass jars with cork stoppers, which he sealed with wax and then boiled. Appert's invention was later improved by others, such as Louis Pasteur, who showed that the process of heating kills microorganisms that cause food to spoil. Today, canning is a common method used to preserve food, and it is used to produce a variety of food products, including fruits, vegetables, meats, and fish.

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Answer:

D. formation of zinc from zinc sulphide ​

Explanation:

A redox reaction could be explained as an artificial reaction in which electrons are moved between two reactants partaking in it. This substitution of electrons can be recognized by examining the variations in the oxidation states of the reacting classes. The generation of hydrogen fluoride is an illustration of a redox reaction. We can crack the reaction down to investigate the oxidation and loss of reactants.

Answer:

0.715 g

Explanation:

mass = 0.0013x550= 0.715 g

12.308% of the chemical is sulfur. The compound's whole sulfur content is transformed into barium sulfate. Therefore, the mass for sulfur from barium sulfate will match the mass for sulfur in the compound.

People are not highly poisonous to sulfur. However, consuming far to much sulfur may result in diarrhea or a burning feeling. Sulfur dust inhalation might irritate the airway or make one cough. The eyes and skin may also become irritated by it.

Sulfur can indeed be found in sulfide ores and in nature. It is native to areas close to volcanoes and hot springs. The tenth most common element, sulfur, is found in almost all types of plant and animal life, meteorites, ocean ocean, the humankind's crust, and or the atmosphere.

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The stable nuclei can be identified if the count of proton and neutron are same then nuclei will be stable.

That kind of nuclei which posses identical number of proton and neutron as well can be referred as stable nuclei. For example: The isotope of carbon \(C^{12}\) has 6 proton and 6 neutron as well. Hence it might be considered as stable nuclei.

A hydrogen atom's nucleus contains positively charged protons. Protons, electrons, as well as neutrons, make up an atom.

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Pre-Lab Report: Trends of Chlorides Across Period 3 Elements

Introduction:

The periodic table is a valuable tool for understanding the properties and trends of elements. One important trend to explore is the variation in properties of chlorides across the period 3 elements. Chlorides are compounds formed by the combination of chlorine with other elements. In this pre-lab experiment, we will investigate the trends in the properties of chlorides across the period 3 elements.

Objective:

The objective of this experiment is to observe and analyze the trends in the properties of chlorides across the period 3 elements in the periodic table.

Materials:

Periodic table

Safety goggles

Laboratory notebook

Pen or pencil

Procedure:

Familiarize yourself with the periodic table and the arrangement of elements in period 3.

Observe the elements in period 3 and identify their corresponding chlorides.

Note down the formulas of the chlorides for each element in period 3.

Analyze the properties of the chlorides, including their physical states (solid, liquid, or gas) at room temperature, solubility in water, and electrical conductivity.

Record your observations and any patterns or trends you notice in the properties of the chlorides.

Safety Considerations:

Wear safety goggles to protect your eyes from any potential hazards.

Follow proper laboratory safety guidelines and procedures.

Data Analysis:

Once you have completed the experiment and recorded your observations, analyze the data to identify any trends in the properties of chlorides across the period 3 elements. Look for patterns in the physical states, solubility, and electrical conductivity of the chlorides.

Discussion and Conclusion:

Based on your observations and data analysis, discuss the trends you observed in the properties of chlorides across the period 3 elements. Explain any patterns you identified and relate them to the underlying principles of periodicity and atomic structure.

In conclusion, this pre-lab experiment aims to investigate the trends in the properties of chlorides across the period 3 elements. By conducting this experiment and analyzing the data, we can deepen our understanding of the periodic table and the variations in chemical properties across different elements.

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Answer:

Oh thats great..

But I hate juice :(

Answer:

D)

Explanation:

CH4 + 2O2 = CO2 + 2H2O

1 (L CO2)= 32.5(L CO2)

1 (CH4) = 32.5 (L CH4)

Answer:

Please give me Brainliest it would mean a lot

One mole of Mg contains  6.022 × 10²³ number of Mg atoms. Hence, number  of moles of Mg that contains 2.5 × 10²⁵ atoms is 41.5 moles.

Any substance  that contains   6.022 × 10²³ number of atoms is called ne mole of that substance. This number is called Avogadro number. Hence, one mole of every elements contains Avogadro number of atoms.

Magnesium is 12th element in periodic table. It is an alkaline earth metal. The number of atoms in one mole of Mg is equal to Avogadro number. Mass of the element with one mole of amount is called its atomic mass.

Number of moles of Mg that contains 2.5 × 10²⁵ atoms :

= 2.5 × 10²⁵ atoms/6.022 × 10²³ = 41.5 moles.

Therefore, 41.5 moles of Mg contains 2.5 × 10²⁵ atoms.

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Answer:

The molecules are held closely together in a regular pattern called a lattice. If the ice is heated, the energy of the molecules increases. ... This is why liquid water is able to flow: the molecules have greater freedom to move than they had in the solid lattice.

Explanation:

I hope this helps and pls mark me brainliest :)

In 100 mg of caffeine, there are 0.002056 moles of nitrogen.

Moles can be defined as a unit of measurement for the amount of a substance. It represents the amount of a chemical substance that contains as many entities such as atoms, molecules, ions, or any other elementary entities as there are in 12 grams of carbon-12.

In one mole of any compound, there are 6.02 × 1023 particles. This number is known as Avogadro's number.

The molecular formula of caffeine is C8H10N4O2. We need to find out the number of moles of nitrogen in 100 mg of caffeine. We can do this by using the following steps.

Step 1: Calculate the molar mass of caffeine.

Molar mass of C8H10N4O2 = 194.19 g/mol.

Step 2: Find the mass of nitrogen in one mole of caffeine.

Mass of nitrogen in caffeine = 4 × (14.01 g/mol) = 56.04 g/mol.

Step 3: Calculate the number of moles of nitrogen in 100 mg of caffeine.

Mass of 100 mg caffeine = 0.1 g.

Number of moles of caffeine = mass ÷ molar mass = 0.1 g ÷ 194.19 g/mol = 0.000514 moles of caffeine

Number of moles of nitrogen in 100 mg of caffeine = number of moles of caffeine × 4

= 0.000514 × 4= 0.002056 moles of nitrogen.

Therefore, there are 0.002056 moles of nitrogen in 100 mg of caffeine.

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An electron domain consists of ; All of the above options  i.e.  a ) nonbonding pair of electrons   b) a single bond     c) a multiple bond.

An Electron domain is the number of bonded locations ( lone pair or bonded ) found around an atom. and the bond location is independent of the type of bonding existing between atoms in a molecule.

Electron domain are also known as electron groups. the types of electron domain are ; nonbonding pair of electrons and bonded atom.

Hence we can conclude that An electron domain consists of ; nonbonding pair of electrons,  a single bond and  a multiple bond.

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Answer:

Explanation:So if a neutron or two is added or removed from the nucleus, then the ... For example Cl–35 has an atomic mass of 35 u (17 protons and 18 neutrons), while Cl–37 ... Calculate the mass contribution of chlorine–35 to the average relative atomic mass ... Which of the following are isotopes of 3517Cl? 1735Cl. 3517Cl. 3717Cl.

Answer: 4 ATP molecules is produced in the glycolysis of 2 glucose molecules

The first stage of most carbohydrate catabolism, which is described as the breakdown of larger molecules into smaller ones, is glycolysis, which is a part of cellular respiration. Two Greek words that mean "to break down anything sweet" are the source of the word glycolysis. Two molecules of ATP are created during the breakdown of glucose and formation of pyruvate during glycolysis.

Explanation:

2 ATP are produced.

For every glucose molecule, meaning that for every 2 glucose molecules, 4 ATP will be created.

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The answer is option

The reaction:

is a Synthesis reaction, because from 2 different substances it is produced

The question requires us to calculate the mass, in grams, contained in 5.26 x 10^-4 mol of HC2H3O2.

To solve this question, first we need to calculate the molar mass of the compound, considering the number of atoms of each element, and then relate the value obtained with the number of moles given (5.26 x 10^-4 mol).

First, to calculate the molar mass of the compound, let's consider the following atomic masses:

atomic mass of C = 12.01 u

atomic mass of H = 1.007 u

atomic mass of O = 15.99 u

Next, we calculate the molar mass. To do that, we need to consider the number of atoms of each element: according to the chemical formula, there are 2 atoms of C, 4 atoms of H and 2 atoms of O:

molar mass (C2H4O2) = (2 * 12.01) + (4 * 1.007) + (2 * 15.99) = 60.03 g/mol

Now, we know that there are 60.03 g for each mol of the compound. With that information, we can estabilish the following relation to calculate the mass contained in 5.26 x 10^-4 mol of the compound:

1 mol --------------------- 60.03 g

5.26 x 10^-4 mol ----- x

Solving for x, we have:

Therefore, there are 0.03158 g of HC2H3O2 in 5.26 x 10^-4 mol of this compound.

The concentration of [\(OH^-\)] is approximately \(3.16 \times 10^{(-11)\) M, and the concentration of [\(H_3O^+\)] is also approximately \(3.16 \times 10^{(-11)\) M.

Based on the given reaction:

\(\[\text{{HP042}} (\text{{aq}}) + \text{{H2O}} (\text{{l}}) \rightleftharpoons \text{{Por}} (\text{{aq}}) + \text{{OH}}^- (\text{{aq}})\]\)

To determine if the equilibrium lies to the left or to the right, we need to examine the relative strengths of the conjugate acid-base pairs involved. Looking at Figure 16.4, we can compare the acidities of the species in the reaction:

\(\(\text{{HP042}}\)\) is a polyprotic acid (phosphoric acid) and can donate three protons (H+ ions).

Por (the conjugate base of \(\(\text{{HP042}}\)\)) is formed when \(\(\text{{HP042}}\)\) donates one proton.

\(H_2O\) is amphiprotic and can act as both an acid and a base.

ΟΗ' (hydroxide ion) is a strong base.

Based on the information from Figure 16.4, we can conclude that:

\(\(\text{{HP042}}\)\) is a weak acid since it donates protons less readily than the strong acids listed in the figure.

Por is a weak base since it accepts protons less readily than the strong bases listed in the figure.

\(H_2O\) is a weaker acid than \(\(\text{{HP042}}\)\) but a stronger base than Por.

ΟΗ' is a strong base.

Now, let's proceed to calculate the concentrations of [\(OH^-\)] and [\(H_3O^+\)] when the pOH is 10.50.

Given: pOH = 10.50

To calculate [\(OH^-\)], we can use the formula:

pOH = -log[\(OH^-\)]

Rearranging the formula, we get:

\([OH^-] = 10^{(-pOH)\)

\([OH^-] = 10^{(-10.50)\)

\(\([OH^-] \approx 3.16 \times 10^{-11} \, \text{M}\)\)

Since the reaction involves the transfer of protons, we know that [\(H_3O^+\)] = [\(OH^-\)] (according to the principle of neutralization for a strong acid and strong base). Therefore:

\(\([H_3O^+] \approx 3.16 \times 10^{-11} \, \text{M}\)\)

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Answer:

Explanation:

An Observation

Molar expansion accounts for the changes in the number of moles of a component within a reactor and can have a significant impact on the volume of a plug flow reactor, particularly in cases where the reactor volume is not constant.

(iii) "ei" is commonly used to represent the molar expansion term in chemical reaction engineering equations. It refers to the change in molar flow rate of a particular component i per unit time, per unit volume. The molar expansion term accounts for the variation in the number of moles of a component within a reactor due to chemical reactions or phase changes.

In a plug flow reactor, molar expansion can have different impacts on the reactor volume depending on the specific case. In part (i), where there is a constant volume, the molar expansion does not affect the reactor volume. The molar flow rates of reactants and products may change due to reactions, but the overall volume remains constant.

In part (ii), where the reactor is a semi-batch reactor with a varying volume, the molar expansion can significantly influence the volume of the reactor. As the reaction proceeds, the molar flow rates of reactants and products change, which can lead to changes in the total number of moles and, consequently, impact the reactor volume. The volume may increase or decrease depending on the molar expansion and the specific reaction taking place.

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Answer:

B. Preliminary conclusion

Explanation:

The answer is B because a hypothesis is really an educated guess you make in regards to a question, usually before performing an experiment, so it can't be a preliminary conclusion because you don't really have any solid evidence to support your hypothesis, you only have background information. You will either be able/not be able to support your hypothesis after your expirement has concluded.

GHS promotes global understanding and facilitates the safe handling, storage, and transportation of chemicals, protecting human health and the environment.

The Globally Harmonized System (GHS) was developed by the United Nations to address the need for a globally consistent approach to classifying and labeling chemicals. It provides a standardized system for identifying and communicating the hazards associated with chemicals, ensuring that this information is easily understood and recognized across different countries and regions.

GHS uses a set of criteria to classify chemicals based on their intrinsic properties and potential hazards. These hazards include physical, health, and environmental effects. By assigning standardized hazard classes, categories, and pictograms, GHS enables individuals and organizations to quickly recognize and assess the dangers posed by different chemicals.

In addition to classification, GHS also emphasizes the importance of clear labeling and safety data sheets (SDS). Chemical containers must display GHS-compliant labels that include hazard pictograms, signal words, hazard statements, and precautionary statements. These labels help users identify the potential risks associated with a chemical and take appropriate safety measures.

Safety data sheets (SDS), formerly known as material safety data sheets (MSDS), provide detailed information about a chemical's hazards, handling precautions, and emergency response measures. GHS establishes a standardized format for SDS, ensuring that essential information is readily available to workers, emergency responders, and others involved in the handling and transport of chemicals.

The adoption of GHS by different countries and organizations around the world has numerous benefits. It enhances the protection of human health and the environment by promoting consistent hazard communication and facilitating risk management practices. GHS enables improved understanding of chemical hazards, leading to safer handling, storage, and transportation of chemicals. It also supports international trade by providing a common language for hazard communication, reducing barriers and enhancing regulatory compliance.

In conclusion, the Globally Harmonized System (GHS) of classification and labeling of chemicals is an international approach to hazard communication. It standardizes the identification, classification, and labeling of chemical hazards, promoting global understanding and ensuring the safe handling, storage, and transportation of chemicals. By adopting GHS, countries and organizations can enhance safety, protect human health and the environment, and facilitate international trade through consistent and effective communication of chemical hazards.

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If the number of protons for a neutral atom is known, the atomic number and the identity of the element can be determined. The atomic number corresponds to the number of protons in an atom, and each element has a unique atomic number.

However, the number of electrons and the number of neutrons cannot be directly identified solely based on the number of protons.

The number of electrons in a neutral atom is equal to the number of protons, as the atom overall carries no net charge. For example, if an atom has 6 protons (atomic number of 6), it will also have 6 electrons to maintain neutrality.

On the other hand, the number of neutrons in an atom can vary, even within the same element. Different isotopes of an element have the same number of protons but different numbers of neutrons. The total mass of an atom, which includes protons and neutrons, is determined by the atomic mass number. However, knowing only the number of protons does not provide information about the specific isotope or the number of neutrons.

To determine the number of neutrons or the specific isotope of an element, additional information such as the mass number or isotopic composition is needed.

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The molar mass (denoted by symbol M) of an element is the mass of one mole of the element. The abbreviation for mole is 'mol'. The molar mass has units of g/mol ('grams per mole'). The molar mass of a compound, or molecule, is the sum of all the molar masses of each individual element that makes up the compound/molecule. Molar mass can be found on a standard IUPAC Periodic Table.

Percentage composition is the percentage mass of each element in the total mass of substance.

\(\large \textsf{M($\rm C_{\,6}H_{\,12}O_{\,6}$) = $12.01\times 6+1.008\times 12+16.00\times 6$}\\\\ \large \textsf{\phantom{M($\rm C_{\,6}H_{\,12}O_{\,6}$)} = $180.156 \rm \ g/mol$}\\\\\)

\(\normalsize \textsf{Hence, percentage composition = molar mass of element $\div$ total molar mass:}\\\\\large \textsf{$\therefore \% \rm \ carbon = \frac{6\times12.01}{180.156}\times100$}\\\\\large \textsf{$\phantom{\therefore \% \rm \ carbon}\approx 40\, \%$}\\\\\)

\(\large \textsf{$\therefore \% \rm \ hydrogen = \frac{12\times1.008}{180.156}\times100$}\\\\\large \textsf{$\phantom{\therefore \% \rm \ hydrogen}\approx 6.7\, \%$}\\\\\)

\(\large \textsf{$\therefore \% \rm \ oxygen = \frac{6\times16.00}{180.156}\times100$}\\\\\large \textsf{$\phantom{\therefore \% \rm \ hydrogen}\approx 53.3\, \%$}\\\\\)

\(\normalsize \textsf{Now applying this technique to the other questions:}\\\\\large \textsf{M($\rm NaCl$) = $22.99 +35.45$}\\\\ \large \textsf{\phantom{M($\rm NaCl$)} = $58.44 \rm \ g/mol$}\\\\\)

\(\large \textsf{$\therefore \% \rm \ sodium = \frac{22.99}{58.44} \times 100$}\\\\\large \textsf{$\phantom{\therefore \% \rm \ sodium}\approx 39.3\, \%$}\\\\\)

\(\large \textsf{$\therefore \% \rm \ chlorine = \frac{35.45}{58.44} \times 100$}\\\\\large \textsf{$\phantom{\therefore \% \rm \ chlorine}\approx 60.7\, \%$}\\\\\)

\(\normalsize \textsf{And finally:}\)

\(\large \textsf{M($\rm Fe_{\,2}(SO_4)_3$) = $55.85\times2+(32.07+16.00\times4)\times3$}\\\\ \large \textsf{\phantom{M($\rm Fe_{\,2}(SO_4)_3$)} = $399.91 \rm \ g/mol$}\\\\\)

\(\large \textsf{$\therefore \% \rm \ iron = \frac{2\times55.85}{399.91} \times 100$}\\\\\large \textsf{$\phantom{\therefore \% \rm \ iron}\approx 27.9\, \%$}\\\\\)

\(\large \textsf{$\therefore \% \rm \ sulfur = \frac{32.07\times 3}{399.91} \times 100$}\\\\\large \textsf{$\phantom{\therefore \% \rm \ sulfur}\approx 24.1\, \%$}\\\\\)

\(\large \textsf{$\therefore \% \rm \ oxygen = \frac{4\times16.00\times3}{399.91} \times 100$}\\\\\large \textsf{$\phantom{\therefore \% \rm \ oxygen}\approx 48.0\, \%$}\\\\\)

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Answer: The thickness of the foil is 0.0516 mm.

To find this, we can use the formula for finding the volume of a rectangular prism, which is V = l x w x h. We are given the volume (V = 0.645 mm3) and the length (l = 10.0 mm) and width (w = 12.5 mm). Rearranging the formula gives us h = 0.645 mm3 / (10.0 mm x 12.5 mm) = 0.0516 mm.

Therefore, the thickness of the foil is 0.0516 mm. This answer was selected from the group of answer choices provided in the question (0.000 516 mm, 80.6 mm, 0.005 16 mm, 0.0516 mm, and 194 mm).


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Answer:

2,2- dichloro-heptane

Explanation:

The IUPAC name of the compound is 2,2-dichloro-heptane

Answer: A

Explanation:

Answer:

A. HE

Explanation:

Answer:

i think this is correct

Explanation:

For amplitude measurements, you count the number of vertical divisions occupied by the signal's vertical signal and then multiply by the vertical scale. The example below shows the process for measuring the peak-to-peak voltage of a signal.