Answer:
first second and last is mechanical while third and fourth is chemical
The volume of titanium with mass of 0. 10g and density of 4. 51 g/cm³ is 0. 02 cm³
<h3>
What is volume?</h3>
Volume is known to be equal to the mass divided by the density.
It is written thus:
Volume = Mass / density
<h3>
How to calculate the volume</h3>
The volume is calculated using the formula:
Volume = mass ÷ density
Given the mass = 0. 10g
Density = 4.51 g/cm³
Substitute the values into the formula
Volume of titanium = 0. 10 ÷ 4.51 = 0. 02 cm³
Thus, the volume of titanium with mass of 0. 10g and density of 4. 51 g/cm³ is 0. 02 cm³
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Newton's first law of motion predicts the behavior of objects where all existing forces are balanced. Objects at equilibrium will not accelerate. Also, an object will only accelerate if there is a net or unbalanced force acting upon it. The presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed and direction.
Newton's second law of motion: behavior of objects for and all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
Answer:
The correct answer will be " RbF > RbCl > RbBr > Rbl".
Explanation:
The size of the given ions will be:
<u>RbCl:</u>
⇒ 689kJ/mol
<u>RbBr:</u>
⇒ 660kJ/mol
<u>Rbl:</u>
⇒ 630kJ/mol
<u>RbF:</u>
⇒ 785kJ/mol
Now according to the size, the arrangement will be:
⇒ (785kJ/mol) > (689kJ/mol) > (660kJ/mol) >(630kJ/mol)
⇒ RbF > RbCl > RbBr > Rbl
The bond among all opposite charging ions seems to be strongest whenever the ions were indeed small.
Answer:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
Explanation:
Let's consider the notation of a galvanic cell.
Cr(s) | Cr³⁺(aq) || Ag⁺(aq) | Ag(s)
On the left, it is represented the anode (oxidation) and on the right, it is represented the cathode (reduction).
The half-reactions are:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
To have the global reaction, we have to multiply the reduction by 3 (so the number of electrons gained and lost are the same) and add both half-reactions.
Global reaction: Cr(s) + 3 Ag⁺(aq) ⇒ Cr³⁺(aq) + 3 Ag(s)
