From the information given, the mass of the hammer seems to be 1.36 kg and the velocity by which it is wielded seems to be 22 m/s.
The most suitable formula to be used here is :
f = m × v
where f is force, m is mass and v is velocity.
f = m × v → f = 1.36 × 22 → f = 29.92 kg m/s
Therefore the force generated by the hammer is 29.92 N
( kg m/s is the same as Newton)
Answer is: freezing point is -0,226°C.
Answer is: the molal concentration of glucose in this solution is 1,478 m.
m(KCl) = 15 g.
n(KCl) = m(KCl) ÷ M(KCl).
n(KCl) = 15 g ÷ 74,55 g/mol.
n(KCl) = 0,2 mol
m(H₂O) = 1650 g ÷ 1000 g/kg = 1,65 kg.
b = n(KCl) ÷ m(H₂O).
b = 0,2 mol ÷ 1,65 kg = 0,122 m.
Kf(water) = 1,86°C/m.
ΔT = Kf(water) · b(solution).
ΔT = 1,86°C/m · 0,122 m.
ΔT = 0,226°C.
Answer:
See explanation
Explanation:
Q1:
Chlorine has 17 protons while magnesium has only 12 protons. Recall that the Zeff depends on the size of the nuclear charge. The greater the size of the nuclear charge, the larger the Zeff experienced by a valence electron.
Q2:
The larger the Zeff, the smaller the atomic radius. Since the valence electrons of Cl experience a greater Zeff than those of Mg due to greater size of the nuclear charge, the atomic radius of chlorine will be smaller than that of Mg.
Q3:
The radius of an atom increases as the value of the principal quantum number (n) increases down the group due to addition of more shells. The greater the number of shells added, the greater the principal quantum number (n) and the greater the atomic radius, hence the answer.
Answer: Oxygen and glucose are both reactants in the process of cellular respiration.
Explanation: Oxygen and glucose are both reactants in the process of cellular respiration. The main product of cellular respiration is ATP; waste products include carbon dioxide and water.
Answer:
The best reagents that are used for the conversion of ethylbenzene to (2-bromoethyl)benzene is shown in the first diagram attached.
Explanation:
Concepts and reason
The concept used to solve this problem is by using the given reagents, possible products will be formed in each step and then label it exactly in the given boxes in order to form the exact product.
Here, the starting reactant is ethyl benzene and the final product is (2-bromoethyl)benzene.
Fundamentals
Bromine molecule is used for bromination of alkene. Trans addition takes place.
Addition of HBr to the double bond forms an alkyl bromide.
Potassium tertiary butoxide is a sterically hindered base.
Bromination of alkane in the presence of sunlight gives radical substitution.
NBS (N-bromosuccinimide) is used for the allylic bromination.
The reaction is as shown in the second attachment(pictures 2,3 and 4).