Combining carbon and sulfur can result in compounds with various physical states at room temperature, such as solids (e.g., carbon disulfide) or gases (e.g., carbon dioxide). The state depends on the specific compound formed.
The type of chemical bonds and interactions that exist between a compound's components, as well as the temperature and pressure levels, all affect the compound's physical state.
While in their pure elemental forms, carbon and sulfur are both solid at ambient temperature, combining them to create a compound can produce a substance with a variety of qualities.
For instance, carbon and sulfur can combine to generate molecules like carbon disulfide, a volatile and extremely combustible liquid that exists at room temperature and at normal atmospheric pressure.
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Explanation:
Conductors and insulators are two types of materials that differ in their ability to conduct electrical current. Here are the key differences between conductors and insulators:
**Conductors:**
1. **Electrical Conductivity:** Conductors are materials that have high electrical conductivity. This means they allow the easy flow of electric charge, typically in the form of electrons. Metals like copper, aluminum, and silver are excellent electrical conductors.
2. **Free Electrons:** Conductors have a large number of free electrons that are loosely bound to their atomic structure. These free electrons can move easily when a voltage is applied, creating an electric current.
3. **Low Resistance:** Conductors have low electrical resistance, which means they offer little opposition to the flow of electric current.
4. **Used for Wiring:** Conductors are commonly used in electrical wiring and circuits to transmit electricity efficiently.
**Insulators:**
1. **Electrical Insulation:** Insulators are materials that have low electrical conductivity. They do not allow the easy flow of electric charge. Examples of insulators include rubber, glass, plastic, and wood.
2. **Tightly Bound Electrons:** Insulators have tightly bound electrons in their atomic structure. These electrons are not free to move, so they do not contribute to the flow of electric current.
3. **High Resistance:** Insulators have high electrical resistance, which means they strongly resist the flow of electric current.
4. **Used for Insulation:** Insulators are used to isolate and protect conductors in electrical systems. They are employed as insulating materials in wires, cables, and electrical devices to prevent electrical leakage and ensure safety.
In summary, conductors allow electric current to flow easily due to the presence of free electrons and low resistance, while insulators inhibit the flow of electric current because their tightly bound electrons and high resistance prevent the movement of charge. These differences make conductors suitable for carrying electricity, while insulators are used to insulate and protect conductive materials in electrical systems.
Answer:
One conclusion from Rutherford's experiment that greatly contradicted Thompson's experiment is that 'atoms are mostly empty space, with positively charged nucleus'.
Explanation:
2Na + 2H2O 2NaOH + H2
Answer: YES
Explanation: The given chemical equation can be balanced by balancing the stiochiometric number of each and every element.
This rule is in accordance with the Law of Conservation of Masses which states that the masses of the elements in any chemical equation are constant at any instant of time.
Thus the given equation can be written as
There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. The mass of NaOH needed for the solution is calculated as follows:
2 M = amount in moles of solute / 2.5 L
amount in moles of solute = 5 mol NaOH
amount in grams = 5 x 40 = 200 grams NaOH
Mass = 200 g
Molarity is the amount of solute dissolved per unit volume of solution. It is expressed as,
Molarity = Moles / Volume of Solution ----- (1)
Data Given;
Molarity = 2.0 mol.L⁻¹
Volume = 2.5 L
M.Mass = 40.0 g.mol⁻¹
First calculate Moles as,
Moles = Molarity × Volume
Putting Values,
Moles = 2.0 mol.L⁻¹ × 2.5 L
Moles = 5 mol
Secondly, calculate Mass using following formula,
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting Values,
Mass = 5 mol × 40.0 g.mol⁻¹
Mass = 200 g
Add 200 g of NaOH in a Volumetric Flask and add distilled water upto the mark of 2.5 L. Shake it and you have prepared a 2.5 L of 2.0 M solution of NaOH.