The volume of the stock solution that has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L is 0.065L.
<h3>How to calculate volume?</h3>
The concentration of a solution can be calculated using the following formula:
C1V1 = C2V2
Where;
- C1 = initial concentration = 1.5M
- C2 = final concentration = 0.54M
- V1 = initial volume = ?
- V2 = final volume = 0.18L
1.5 × V1 = 0.54 × 0.18
1.5V1 = 0.0972
V1 = 0.0972 ÷ 1.5
V1 = 0.065L
Therefore, the volume of the stock solution that has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L is 0.065L.
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Answer:
All description is given in explanation.
Explanation:
Van der Waals forces:
It is the general term used to describe the attraction or repulsion between the molecules. Vander waals force consist of two types of forces:
1. London dispersion forces
2. Dipole-dipole forces
1. London dispersion forces:
These are the weakest intermolecular forces. These are the temporary because when the electrons of atoms come close together they create temporary dipole, one end of an atom where the electronic density is high is create negative pole while the other becomes positive . These forces are also called induce dipole- induce dipole interaction.
2. Dipole-dipole forces:
These are attractive forces , present between the molecules that are permanently polar. They are present between the positive end of one polar molecules and the negative end of the other polar molecule.
Hydrogen bonding:
It is the electrostatic attraction present between the atoms which are chemically bonded. The one atom is hydrogen while the other electronegative atoms are oxygen, nitrogen or flourine. This is weaker than covalent and ionic bond.
Ionic bond or electrostatic attraction:
It is the electrostatic attraction present between the oppositely charged ions. This is formed when an atom loses its electron and create positive charge and other atom accept its electron and create negative charge.
Hydrophobic interaction:
It is the interaction between the water and hydrophobic material. The hydrophobic materials are long chain carbon containing compound. These or insoluble in water.
Covalent bond:
These compounds are formed by the sharing of electrons between the atoms of same elements are between the different element's atoms. The covalent bond is less stronger than ionic bond so require less energy to break as compared to the energy require to break the ionic bond.
The mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g
<h3>Calculating mass </h3>
From the question we are to calculate the mass of NaHCO₃ (sodium bicarbonate) used in the experiment
From the given information
Mass of empty evaporating dish = 46.233g
Mass of evaporating dish + Sodium bicarbonate = 48.230g
∴ Mass of sodium bicarbonate (NaHCO₃) = [Mass of evaporating dish + Sodium bicarbonate] - [Mass of empty evaporating dish]
Mass of sodium bicarbonate (NaHCO₃) = 48.230g - 46.233g
Mass of sodium bicarbonate (NaHCO₃) = 1.997 g
Hence, the mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g
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Answer:
Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}
Explanation:
First, we have to write the balanced chemical equation for the reaction. Nitrogen monoxide (NO) reacts with water (H₂O) to give ammonia (NH₃) and oxygen (O₂), according to the following:
NO(g) + H₂O(g) → NH₃(g) + O₂(g)
To balance the equation, we add the stoichiometric coefficients (4 for NH₃ and NO to balance N atoms, then 6 for H₂O to balance H atoms and then 5 for O₂ to balance O atoms):
4 NO(g) + 6 H₂O(g) → 4 NH₃(g) + 5 O₂(g)
All reactants and products are in the gaseous phase, so the equilibrium constant is expressed in terms of partial pressures (P) and is denoted as Kp. The Kp is expressed as the product of the reaction products (NH₃ and O₃) raised by their stoichiometric coefficients (4 and 5, respectively) divided into the product of the reaction reagents (NO and H₂O) raised by their stoichiometric coefficients (4 and 6, respectively). So, the pressure equilibrium constant expression is written as follows:
