KE = mv²/2
m=2*KE/v²
v=50 m/s
KE=500J
m=2*KE/v² =2*500/50²=1000/2500= 0.4 kg
Answer:
1. 7.256g of NaCl
2. 47.33g of Cl2
Explanation:
2 moles of Na reacts to produce 2 moles of NaCl
8 moles of Na will still produce 8 moles of NaCl
Mass of NaCl = molar mass of Nacl/moles of Nacl
=58.5/8
=7.256g of NaCl
From the equation, 2 moles of Na reacts with 1 mole of Cl2
3/2 moles of Cl2 will react with 3 moles of Na
Mass of Cl2 = 71/1.5
=47.33g of Cl2
Explanation:
The right answer is 2.
The number of protons contained in a nucleus (called an atomic number) is characteristic of a chemical element. For a given atomic number, the number of neutrons defines different "types" of this element: isotopes. The variation of the number of protons of the nucleus of an atom, during a nuclear reaction for example, causes a change of the element studied.
Answer : Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.
Explanation :
The given molecule is, 
Three types of inter-molecular forces are present in this molecule which are Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.
- Hydrogen-bonding : when the partial positive end of hydrogen is bonded with the partial negative end of another molecule like, oxygen, nitrogen, etc.
- Dipole-dipole attraction : When the partial positively charged part of the molecule is interact with the partial negatively charged part of the molecule. For example : In case of HCl.
- London-dispersion force : This force is present in all type of molecule whether it is a polar or non-polar, ionic or covalent. For example : In case of Br-Br , F-F, etc
Hydrogen-bonding is present between the oxygen and hydrogen molecule.
Dipole-dipole forces is present between the carbon and oxygen molecule.
London-dispersion forces is present between the carbon and carbon molecule.
Answer:
Aluminum iodide (AlI₃)
Explanation:
The synthesis reaction of aluminum (Al) and iodine (I) can be illustrated as shown below:
Aluminium exhibit trivalent positive ion (Al³⁺)
Iodine exhibit univalent negative ion (I¯)
During reaction, there will be an exchange of ion as shown below:
Al³⁺ + I¯ —> AlI₃
Thus, we can write the balanced equation for the reaction as follow:
Al + I₂ —› AlI₃
There are 2 atoms of I on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of AlI₃ and 3 in front of I₂ as shown below:
Al + 3I₂ —› 2AlI₃
There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Al as shown below:
2Al + 3I₂ —› 2AlI₃
Thus the equation is balanced.
The product on the reaction is aluminum iodide (AlI₃)
