Answer: <span>A fewer number of particles of the sample will dissolve in 1 minute.
That is because normally the solubility and rate of solubility of the salts in water increase with the temperature. This is, the higher the temperature the higher and faster the number of particles that the water can dissolve. So, at 70°C more particles will be dissolved in water in 1 minute than at 20°C.
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Answer: The correct answer is "B" two bonding domains(or bonding pairs) or two non bonding domains(or lone pairs)
Explanation:
Molecular geometry/structure is a three dimensional shape of a molecule. It is basically an arrangement of atoms in a molecule.It is determined by the central atom, its surrounding atoms and electron pairs.According to VSEPR theory, there are 5 basic shapes of a molecule: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.
A)Four bonding domains and zero non bonding domains: shape is tetrahedral and bond angle is 109.5°
B)Two bonding domains and two non bonding domains(lone pairs): shape is bent and bond angle is 104.5°
C)Three bonding domains and one non bonding domain: shape is trigonal pyramidal and bond angle is 107°
D)Two bonding domain and zero non bonding domain: shape is linear and bond angle is 107°
E)Two bonding domain and one non bonding domain: bent shape and bond angle is 120°
F)Three bonding domains and zero nonbonding domain: shape is trigonal planar and bond angle is 120°
Hence Two bonding domains and two non bonding domains have the smallest bond angle.
Answer:
The answer is <u>applied research</u>
Explanation:
Pure research becomes <u>applied research</u> when scientists develop a hypothesis based on the data and try to solve a specific problem.
This is because the pure research try to understand, predict or explain the behavior of different phenomena <em>(the data)</em> while the applied research try to develop new technologies or methods (<em>hypothesis)</em> to take part, intervene and/or create changes on these phenomena and solve a <em>specific problem.</em>
NaOH + CH3COOH -> CH3COONa + H20