Answer:
k = 23045 N/m
Explanation:
To find the spring constant, you take into account the maximum elastic potential energy that the spring can support. The kinetic energy of the car must be, at least, equal to elastic potential energy of the spring when it is compressed to its limit. Then, you have:
(1)
M: mass of the car = 1050 kg
k: spring constant = ?
v: velocity of the car = 8 km/h
x: maximum compression of the spring = 1.5 cm = 0.015m
You solve the equation (1) for k. But first you convert the velocity v to m/s:
The spring constant is 23045 N/m
Answer:
pH → 7.46
Explanation:
We begin with the autoionization of water. This equilibrium reaction is:
2H₂O ⇄ H₃O⁺ + OH⁻ Kw = 1×10⁻¹⁴ at 25°C
Kw = [H₃O⁺] . [OH⁻]
We do not consider [H₂O] in the expression for the constant.
[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ M
Kw depends on the temperature
0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻] → [H₃O⁺] = [OH⁻] at 0°C
√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M
- log [H₃O⁺] = pH
pH = - log 3.46×10⁻⁸ → 7.46
Answer:
In 1000 ml there is 0.10 moles of Fe 2+
Therefore, in 10 ml there is (0.1/1000)*10= 0.001 mol of Fe2+
mole ratio for rxn Fe2+ : MnO4- is
1 : 2
therefore if 0.001 moles of Fe2+ react then 0.001*2 =0.002 moles of MnO4- react with Fe2+
hence, molarity of MnO4- = (mol*vol)/1000
= 0.002*10.75/1000= 2.15*10-5M
Explanation:
Hope this helps
Answer: Option (e) is the correct answer.
Explanation:
A bond that is formed when an electron is transferred from one atom to another results in the formation of an ionic bond.
For example, NaBr will be an ionic compound as there is transfer of electron from Na to Br.
Whereas a bond that is formed by sharing of electrons is known as a covalent bond.
For example, 
will be a covalent compound as there is sharing of electron between carbon and bromine atom.
Also, when electrons are shared between the combining atoms and there is large difference in electronegativity of these atoms then partial charges develop on these atoms. As a result, it forms a polar covalent bond.
For example, in a HBr compound there is sharing of electrons between H and Br. Also, due to difference in electronegativity there will be partial positive charge on H and partial negative charge on Br.
Thus, we can conclude that out of the given options HBr is the only compound that has polar covalent bonds.
