When dissolved in water, most Group 1 metal salts can be described as strong electrolytes.
When Group 1 metal salts are dissolved in water, they can be described as strong electrolytes. This is because Group 1 metals, such as lithium (Li), sodium (Na), potassium (K), and so on, readily lose their outermost valence electron to form positive ions (cations). These cations then dissociate completely in water, separating from the anions to which they were originally bonded.
The dissociation of Group 1 metal salts in water results in the formation of positively charged metal ions and negatively charged non-metal ions (anions). These ions are free to move and conduct electric current, making the solution a good conductor of electricity. The complete dissociation of Group 1 metal salts in water and the presence of freely moving ions make them strong electrolytes.
Strong electrolytes are substances that ionize completely or almost completely in solution, producing a high concentration of ions. This is in contrast to weak electrolytes, which only partially ionize and produce a lower concentration of ions.
In summary, when Group 1 metal salts are dissolved in water, they form strong electrolytes due to their ability to dissociate completely into ions, leading to a high concentration of freely moving ions in the solution, thus enabling efficient electrical conductivity.
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Magnesium has three stable isotopes. The most commonly occurring isotope, Mg24,
has an isotopic mass of 23.985 u and makes up 78.99% of naturally occurring magnesium atoms. The isotope Mg25 makes up 10.00% of magnesium atoms and has an isotopic mass of 24.986 u. The isotope Mg26 makes up 11.01%
of magnesium atoms and has an isotopic mass of 25.983 u .
Using the isotopic composition provided, calculate the average atomic mass of magnesium.
average atomic mass:
The average atomic mass of magnesium is 24.305 amu.
The mass of a single atom is too small to be measured directly, but chemists and physicists use the atomic mass unit (amu) to describe the mass of an atom.
The atomic mass unit is defined as 1/12th of the mass of a carbon-12 atom. Magnesium (Mg) has three isotopes: Mg-24, Mg-25, and Mg-26, with masses of 23.985 amu, 24.986 amu, and 25.983 amu, respectively.
These isotopes make up 78.99%, 10.00%, and 11.01%, respectively, of all naturally occurring Mg atoms.
The atomic mass of Mg can be calculated using the weighted average of the three isotopes as follows:Average atomic mass = (78.99/100 x 23.985 amu) + (10.00/100 x 24.986 amu) + (11.01/100 x 25.983 amu)Average atomic mass = 23.940 amu + 2.499 amu + 2.864 amuAverage atomic mass = 29.303 amu.
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Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia.
N2(g)+3H2(g)⟶2NH3(g)
There are four molecules of nitrogen and nine molecules of hydrogen present in the diagram.
When the reaction is complete, how many molecules of NH3 are produced?
What is the limiting reactant?
How many molecules of each reactant are remain after the reaction is complete?
After the reaction is complete, no nitrogen and no hydrogen molecules remain, and 8.00 x 1014 molecules of NH3 are produced.
In the equation, nitrogen and hydrogen react at a high temperature, in the presence of a catalyst, to produce ammonia, according to the balanced chemical equation:N2(g)+3H2(g)⟶2NH3(g)The coefficients of each molecule suggest that one molecule of nitrogen reacts with three molecules of hydrogen to create two molecules of ammonia.
So, to determine how many molecules of ammonia are produced when four nitrogen and nine hydrogen molecules are present, we must first determine which of the two reactants is the limiting reactant.
To find the limiting reactant, the number of moles of each reactant present in the equation must be determined.
Calculations:
Nitrogen (N2) molecules = 4Hence, the number of moles of N2 = 4/6.02 x 1023 mol-1 = 6.64 x 10-24 mol
Hydrogen (H2) molecules = 9Hence, the number of moles of H2 = 9/6.02 x 1023 mol-1 = 1.50 x 10-23 mol
Now we have to calculate the number of moles of NH3 produced when the number of moles of nitrogen and hydrogen are known, i.e., mole ratio of N2 and H2 is 1:3.
The mole ratio of N2 to NH3 is 1:2; thus, for every 1 mole of N2 consumed, 2 moles of NH3 are produced.
The mole ratio of H2 to NH3 is 3:2; thus, for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
From these mole ratios, it can be observed that the limiting reactant is nitrogen.
Calculation for NH3 production:
Nitrogen (N2) moles = 6.64 x 10-24 moles
The mole ratio of N2 to NH3 is 1:2; therefore, moles of NH3 produced is 2 × 6.64 × 10−24 = 1.33 × 10−23 moles.
Now, to determine how many molecules of NH3 are produced, we need to convert moles to molecules.
1 mole = 6.02 x 1023 molecules
Thus, 1.33 x 10-23 moles of NH3 = 8.00 x 1014 molecules of NH3 produced.
To find the amount of each reactant remaining after the reaction is complete, we must first determine how many moles of nitrogen are consumed, then how many moles of hydrogen are consumed, and then subtract these from the initial number of moles of each reactant.
The moles of nitrogen consumed = 4 moles × 1 mole/1 mole N2 × 2 mole NH3/1 mole N2 = 8 moles NH3
The moles of hydrogen consumed = 9 moles × 2 mole NH3/3 mole H2 × 2 mole NH3/1 mole N2 = 4 moles NH3
Thus, the moles of nitrogen remaining = 6.64 × 10−24 mol – 8 × 2/3 × 6.02 × 10^23 mol-1 = 5.06 × 10−24 mol
The moles of hydrogen remaining = 1.50 × 10−23 mol – 4 × 2/3 × 6.02 × 10^23 mol-1 = 8.77 × 10−24 mol
Finally, the number of molecules of each reactant remaining can be calculated as follows:
Number of N2 molecules remaining = 5.06 × 10−24 mol × 6.02 × 10^23 molecules/mol = 3.05 × 10−1 molecules ≈ 0 molecules
Number of H2 molecules remaining = 8.77 × 10−24 mol × 6.02 × 10^23 molecules/mol = 5.28 × 10−1 molecules ≈ 0 molecules.
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Why is it easy to describe things we can see?
Answer:
when you're trying to describe what you see, focus on describing the formal elements of a visual: line, colour, composition, texture, depth, symmetry, size etc. before elaborating on the value of what you want to see.
Why did researchers believe that the particle left after electrons were emitted as cathode rays
had to be positive?
This is found out that the particle left after electrons were emitted as cathode rays is actually a positively charged particle, the proton, which is a subatomic particle present in the nucleus of an atom.
When electrons were emitted as cathode rays, researchers believed that the particle left after electrons were emitted as cathode rays had to be positive because of several reasons. The cathode rays were found to bend towards the positive pole of the electric field or the anode in the cathode ray tube, which means the rays were made up of negatively charged particles that move away from the negatively charged cathode and towards the positively charged anode.Another reason for the belief that the particle left after electrons were emitted as cathode rays had to be positive was the fact that the cathode rays consisted of fast-moving electrons, which, when they collided with the particles in the air inside the cathode ray tube, produced positively charged ions. The positive ions created a positive atmosphere in the tube, which pulled the cathode rays towards the anode.This belief was further strengthened when researchers discovered that cathode rays, which consist of negatively charged particles, can cause the emission of positively charged particles when they collide with matter. This led to the conclusion that the particle left after electrons were emitted as cathode rays had to be positive.For more such questions on positively
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What is the median reaction of second end point in HCL and NaOH titration
The median reaction at the second end point in the HCl and NaOH titration is: HCl + NaOH → NaCl + H2O
In a titration between hydrochloric acid (HCl) and sodium hydroxide (NaOH), the reaction involved is the neutralization reaction between an acid and a base. The balanced equation for this reaction is:
HCl + NaOH → NaCl + H2O
In this reaction, one mole of HCl reacts with one mole of NaOH to form one mole of NaCl (sodium chloride) and one mole of water.
During the titration process, the reaction occurs gradually as the base is added to the acid solution.
The first end point of the titration is reached when the moles of HCl and NaOH are stoichiometrically equivalent, meaning they react in a 1:1 ratio. At this point, all the HCl has been neutralized by the NaOH, and no excess of either reagent remains.
However, if the titration is continued beyond the first end point, the reaction between HCl and NaOH can still occur, albeit in a different ratio.
The second end point refers to the point where the moles of NaOH added exceed the stoichiometrically required amount to neutralize the HCl completely. As a result, any excess NaOH added after the second end point reacts with the excess HCl in a 1:1 ratio.
Therefore, the median reaction at the second end point in the HCl and NaOH titration is:
HCl + NaOH → NaCl + H2O
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a student has a 1 L solution of 2 M HCL and wants to increase the HCL concentration to 3 M
The student needs to add approximately 83.3 mL of 12 M HCl solution to the existing 1 L of 2 M HCl solution to increase the concentration to 3 M. It is important to handle concentrated acids with caution and follow proper safety procedures.
To increase the concentration of a 1 L solution of 2 M HCl to 3 M, the student needs to calculate the volume of concentrated HCl needed and add it to the existing solution. Here's how the calculation can be done:
Given:
Initial concentration of HCl solution = 2 M
Final concentration desired = 3 M
Initial volume of HCl solution = 1 L
Step 1: Calculate the moles of HCl in the initial solution.
Moles of HCl = Initial concentration × Initial volume = 2 M × 1 L = 2 moles
Step 2: Calculate the moles of HCl needed for the desired concentration.
Moles of HCl needed = Final concentration × Final volume = 3 M × 1 L = 3 moles
Step 3: Calculate the moles of HCl to be added.
Moles of HCl to be added = Moles needed - Moles present = 3 moles - 2 moles = 1 mole
Step 4: Convert the moles of HCl to the required volume of concentrated HCl.
To calculate the volume, we need to know the concentration of the concentrated HCl solution. Assuming it is 12 M, we can use the following formula:
Volume of concentrated HCl = Moles of HCl to be added / Concentration of concentrated HCl
Volume of concentrated HCl = 1 mole / 12 M = 0.0833 L or 83.3 mL
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Draw the Lewis structure for a hydrogen (H₂) molecule.
The Lewis structure of the hydrogen molecule have been shown in the image attached.
What is the Lewis structure?The Lewis structure, sometimes referred to as the Lewis dot structure or electron dot structure, is a figurative picture of a molecule or ion that shows how the atoms are arranged, how they are bound together, and how the valence electrons are distributed.
We can see that there are two electrons that have been shared by the hydrogen atoms and it is positioned in between the atoms as shown
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the valency of magnesium is 2 why
The valency of magnesium is 2 because it has two valence electrons in its outermost shell. Magnesium belongs to the second group of the periodic table, which means it has two electrons in its outermost shell or valence shell. In order to achieve a stable configuration, magnesium tends to lose these two valence electrons to form a positively charged ion with a charge of 2+. This makes magnesium a bivalent or divalent element with a valency of 2.
Answer:
Because the outer shell of magnesium contains 2 atoms
In the combustion of hydrogen gas, hydrogen reacts with oxygen from the air to form water vapor. hydrogen+oxygen⟶water
If you burn 46.2g of hydrogen and produce 413g of water, how much oxygen reacted?
mass of oxygen:
Answer:
ok, here is your answer
Explanation:
AI-generated answer
To find the mass of oxygen that reacted, we need to use the Law of Conservation of Mass, which states that in a chemical reaction, the mass of the reactants equals the mass of the products.
First, we need to find the number of moles of hydrogen that reacted:
Molar mass of hydrogen (H₂) = 2.016 g/mol
Number of moles of H₂ = mass/molar mass = 46.2 g/2.016 g/mol = 22.92 mol
Next, we need to use the balanced chemical equation to find the number of moles of water produced:
hydrogen + oxygen → water
2H₂ + O₂ → 2H₂O
From the equation, we can see that for every 2 moles of H₂, 1 mole of O₂ is required to produce 2 moles of H₂O. Therefore, the number of moles of O₂ required to produce 22.92 moles of H₂O is:
Number of moles of O₂ = 1/2 x 22.92 mol = 11.46 mol
Finally, we can find the mass of oxygen that reacted by using its molar mass:
Molar mass of oxygen (O₂) = 32.00 g/mol
Mass of oxygen = number of moles x molar mass = 11.46 mol x 32.00 g/mol = 366.72 g
Therefore, the mass of oxygen that reacted is 366.72 g.
mark me as brainliestBriefly explain the features of computer?
Some of the essential features of computers:
1. Processing Power
2. Storage Capacity
3. Memory
4. Input and Output
5. Connectivity
6. Software
7. Multitasking
8. Scalability
9. Reliability and Durability
10. User Interface
Computers are complex machines that have become an integral part of our daily lives. They possess several key features that enable them to perform a wide range of tasks efficiently. Here are some of the essential features of computers:
1. Processing Power: Computers are capable of executing complex calculations and tasks at incredible speed. They contain powerful processors that can perform billions of operations per second, enabling them to handle various applications and processes.
2. Storage Capacity: Computers have the ability to store and retrieve vast amounts of data. They come equipped with different types of storage devices such as hard drives, solid-state drives (SSDs), and cloud storage, providing ample space for storing files, programs, and operating systems.
3. Memory: Computers have two primary types of memory: RAM (Random Access Memory) and ROM (Read-Only Memory). RAM provides temporary storage for data and instructions that the computer is actively using, while ROM contains firmware and permanent instructions that are essential for booting up the computer.
4. Input and Output: Computers allow users to interact with them through various input and output devices. Input devices include keyboards, mice, touchscreens, and microphones, while output devices include monitors, printers, speakers, and headphones. These devices enable users to input commands and receive feedback from the computer.
5. Connectivity: Computers can connect to networks and other devices, enabling communication and data transfer. They have built-in network adapters and support for various connectivity options such as Ethernet, Wi-Fi, Bluetooth, and USB, allowing users to access the internet, share files, and connect to peripherals.
6. Software: Computers run on software, including operating systems, applications, and utilities. The software provides the instructions and programs that allow users to perform tasks, manipulate data, and manage hardware resources.
7. Multitasking: Computers are designed to handle multiple tasks simultaneously. They can execute multiple programs concurrently, switch between tasks rapidly, and allocate system resources efficiently, providing users with the ability to multitask and enhance productivity.
8. Scalability: Computers offer scalability, allowing users to upgrade hardware components and expand storage capacity as needed. This feature ensures that computers can adapt to evolving technological demands and accommodate future growth.
9. Reliability and Durability: Computers are designed to be reliable and durable. They undergo rigorous testing and are built with high-quality components to ensure stable performance and withstand daily use.
10. User Interface: Computers provide graphical user interfaces (GUIs) that enable users to interact with the system easily. GUIs utilize visual elements such as icons, windows, and menus, making it intuitive for users to navigate and access various functions.
These features collectively make computers versatile, powerful, and indispensable tools in our modern world, enabling us to accomplish a wide range of tasks efficiently and effectively.
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Write the chemical formula for this molecule
A 75.0- mL
volume of 0.200 M
NH3
( Kb=1.8×10−5
) is titrated with 0.500 M
HNO3
. Calculate the pH
after the addition of 17.0 mL
of HNO3
.
Answer:
ok, here is your answer
Explanation:
i am going to solve this problem by using the ICE table method which is an easy method to determine the pH of a weak base with the given data of the problem.Given:Initial volume of NH3 solution (Vi) = 75.0 mLInitial concentration of NH3 solution (Ci) = 0.200 MInitial moles of NH3 solution (Ni) = Ci x Vi = 0.200 M x 75.0 mL = 0.0150 molesKb = 1.8 x 10^-5Moles of HNO3 added (n) = 0.500 M x 17.0 mL = 0.00850 molesVolume of NH3 solution after the addition of HNO3 (Vf) = 75.0 mL + 17.0 mL = 92.0 mLConcentration of NH3 solution after the addition of HNO3 (Cf) = Ni / Vf = 0.0150 moles / 92.0 mL = 0.163 MTo find the pH after the addition of 17.0 mL of HNO3, we need to use the ICE table method.ICE table method:Initial: NH3 + H2O ⇌ NH4+ + OH-Change: -x 0 +x +xEquilibrium: 0.0150 - x 0 x xKb = [NH4+][OH-] / [NH3]1.8 x 10^-5 = x^2 / 0.163Solving for x, x = 0.00171 M[OH-] = 0.00171 M[OH-] = Kw / [H3O+] = 1.0 x 10^-14 / [H3O+][H3O+] = 5.85 x 10^-12pH = -log[H3O+]pH = -log(5.85 x 10^-12)pH = 11.23Therefore, the pH after the addition of 17.0 mL of HNO3 is approximately 11.23.
mark me as brainliestHow did he show that these particles had a charge on them?
J.J. Thomson discovered electrons and their negative charge through the cathode ray experiment, leading to the development of the plum pudding model of the atom.
J.J. Thomson, a British physicist, was the first to discover electrons in 1897.
He conducted the cathode ray experiment to identify the negatively charged particles.
The cathode ray tube is a vacuum-sealed glass tube with two electrodes at each end: a cathode and an anode.
When a high voltage electrical current is applied to the electrodes, the tube glows, indicating that the cathode rays are being emitted from the cathode and traveling through the tube towards the anode.
The cathode rays were found to have a negative charge, according to Thomson.
These rays were identified as particles by the presence of a magnet, which caused the particles to bend in the direction opposite to the magnet's polarity.
This discovery indicated that the particles had a charge on them because they were deflected by the magnetic field, which is only possible if the particles have an electric charge.
Thomson further concluded that these particles were about 1,000 times smaller than hydrogen atoms because of the degree of deflection they experienced in the magnetic field.
Furthermore, Thomson created the plum pudding model of an atom, in which electrons are dispersed throughout a positively charged matrix, based on his findings.
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