Answer:
Cp= 1.005 kJ/kg °C = 1.005 kJ/(kg*K) = 1.005 J/g°C = 4.206 J/g°C = 0.776 BTU/lb°F
Explanation:
for the specific heat
1) Cp= 1.005 kJ/kg °C * (1 °C/ 1 K) (temperature differences) = 1.005 kJ/(kg*K)
2) Cp= 1.005 kJ/kg °C * (1000 J/ kJ)* (1 kg/1000 gr) = 1.005 J/g°C
3) Cp= 1.005 kJ/kg °C * (4.186 kcal/kJ) = 4.206 J/g°C
4) Cp= 1.005 kJ/kg °C * (1 BTU/1.055 kJ)* (0.4535 kg/lb)*(1.8°C/°F)= 0.776 BTU/lb°F
Answer:
Label
Explanation:
You need to search if the chemical need a glass or plastic container. Also if the chemical need to be in a clear or dark container that can protect it from light. Also if need to be a certain temperature, for example, if need to be in the fridge or can be at room temperature.
Most important is that you need to label with the name, formula, date, and if you have the number of lot and the expiration date.
(2) 50 K and 600 kPa (4) 750 K and 600 kPa
Answer : The correct option is, (3) 750 K and 20 kPa
Explanation :
The conditions for ideal gas are :
Ideal gas are those gas that has no intermolecular attractions.
Ideal gas are those gas that have negligible volume.
The ideal gas equation is,
The conditions for real gas are :
Real gas are those gas that have intermolecular attractions.
Real gas are those gas that have volume.
The real gas equation is,
A real gas behave ideally at high temperature and low pressure condition.
From the given options, option (3) have high temperature and low pressure is the correct option.
Hence, at 750 K and 20 kPa conditions of temperature and pressure does a sample of helium behave most like an ideal gas.
Using Einstein’s famous equation, E = mc^2
Given mass, m = 0.112 kg and knowing c = 3.0 x 10^8 m/s, energy E can be determined
E = (0.112 kg)(3.0 x 10^8 m/s)^2
= 1.008 x 10^16 J
Hope this helps
Answer:
Joules is the energy equivalent of this .
Explanation:
Mass of hydrogen atoms in 1 kg of water = 112 g= 0.112 kg
Mass and energy related together and that expression was given by the famous scientist named Albert Einstein:
E = Energy related to mass m
c = Speed of the light
m = 0.112 kg , c =
Joules is the energy equivalent of this .
Answer:
Explanation:
Substances with giant covalent structures are solids with high melting and boiling points due to the nature of the covalent bonds and the three-dimensional network they form within the crystal lattice. This structure is also often referred to as a network covalent structure. Let's break down the key reasons why these substances have such properties:
1. **Strong Covalent Bonds**: In giant covalent structures, each atom forms strong covalent bonds with neighboring atoms. Covalent bonds involve the sharing of electrons between atoms. This sharing results in the formation of very strong and directional bonds, which require a significant amount of energy to break.
2. **Three-Dimensional Network**: In these substances, the covalent bonds extend in a three-dimensional network throughout the entire structure. This means that every atom is bonded to several neighboring atoms in all three spatial dimensions. This extensive network of covalent bonds creates a robust and interconnected structure.
3. **Lack of Weak Intermolecular Forces**: Unlike some other types of solids (e.g., molecular solids or ionic solids), giant covalent structures lack weak intermolecular forces, such as Van der Waals forces. In molecular solids, weak intermolecular forces are responsible for their relatively low melting and boiling points. In giant covalent structures, the primary forces holding the atoms together are the covalent bonds themselves, which are much stronger.
4. **High Bond Energy**: The covalent bonds in giant covalent structures have high bond energies, meaning that a substantial amount of energy is required to break these bonds. When a solid is heated, the energy provided must be sufficient to overcome the covalent bonds' strength, leading to the high melting and boiling points.
5. **Rigidity and Structural Integrity**: The three-dimensional covalent network imparts rigidity and structural integrity to the substance. This network resists deformation and allows the substance to maintain its solid form at high temperatures, as the covalent bonds continuously hold the structure together.
Examples of substances with giant covalent structures include diamond (composed of carbon atoms), graphite (also composed of carbon atoms but arranged differently), and various forms of silica (e.g., quartz and silicon dioxide). Diamond, in particular, is known for its exceptional hardness, high melting point, and remarkable optical properties, all of which are attributed to its giant covalent structure.
In summary, giant covalent structures have high melting and boiling points because of the strong covalent bonds, the three-dimensional network of bonds, and the absence of weak intermolecular forces. These factors combine to create a solid with exceptional stability and resistance to temperature-induced phase changes.
Substances with simple molecular structures are usually gases, liquids, or solids with low melting points due to the intermolecular forces between their molecules. The chemical identities of the molecules determine the types and strengths of these attractions, influencing the physical state of the substance.
Substances with simple molecular structures tend to be gases, liquids, or solids with low melting and boiling points because of the nature of intermolecular forces at play. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. For instance, small, symmetrical molecules, such as H2, N2, O2, and F2, have weak intermolecular attractive forces and form molecular solids with very low melting points (below -200 °C).
In a liquid, intermolecular attractive forces hold the molecules together, though they still have sufficient kinetic energy to move relative to each other. In gases, the molecules have large separations compared to their sizes due to which the forces between them can be ignored, except during collisions.
Therefore, the chemical identities of the molecules in a substance determine the types and strengths of intermolecular attractions possible; this subsequently influences whether the substance is a gas, liquid, or solid, and its melting and boiling points.
Learn more about Intermolecular Forces here:
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