Answer: Hydrogen bonds
Explanation: Hydrogen bonds allow two molecules to link together temporarily. Water molecules are made up of two hydrogen atoms and one oxygen atom, held together by polar covalent bonds.
Lifting the backpack off the floor. Force is being applied in only one direction then (up) which is what constitutes as work. Carrying the box of crayons applies force in two directions (up and forward), which cancel each other out. Work has a vector, which is a quantity containing both direction and magnitude (one, finite direction, not two).
Given data Atomic mass of Ra= 226g/mol
no. of moles =1.0/226g/mol =0.04424moles
no. of atoms in 0.044moles
no. of atoms =no. of moles x avogadro's number
= 0.044x 6.022 x10^23 = 0.264968 x 10^22
If 10^15 atoms of Ra produce 1,373*10^4 atoms of<u> Rn per second</u> then 2,66 *10^21 forms 3,658*10^10 atoms of Rn per second.
Day has 246060=86400 s
That means that 2,66x10^21 atoms of Ra produces 3,16 x10^15 atoms of Rn in a day.
N(Rn)=3.16* 10 ^15 n(Rn)=N/NA
n(Rn)=5,25*10−9 pV=nR*T
T=273.15K R=8,314
p=101325Pa V=n∗R∗T/p
V=5.25∗10^−9 ∗ 8.314 ∗ 273.15 / 101325
V=1.1810^−10 m^3 = 118 x10^-7 liters of Rn, measured at STP, are produced per day by 1.0 g of Ra
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This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.
Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:
Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:
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(a) The displacement of point P at time t=0.10s is determined as +2cm.
(b) The displacement of point P at time t=0.20s is determined as -2cm.
<h3>
What is displacement?</h3>
Displacement is the change in position of an object. It is obtained from the product of velocity and time of motion.
x = vt
<h3>Displacement of the waves after 0.1 s</h3>
x = 10 m/s x 0.1 s = 1 m
Each wave will travel 1 m to the right or to the left, depending on the initial direction.
- wave B from left will stop at point 0 m
- wave A from left will stop at point -1 m
- wave C from right will stop at point 0 m
- wave D from right will stop at point + 1 m
wave B and C superimposed and the displacement will be between A and D.
Amplitude of A = - 2cm
Amplitude of D = + 4cm
Displacement of point P = 4 cm - 2 cm = 2cm
<h3>Displacement of the waves after 0.2 s</h3>
x = 10 m/s x 0.1 s = 2 m
Each wave will travel 2 m to the right or to the left, depending on the initial direction.
- wave B from left will stop at point 1 m
- wave A from left will stop at point 0 m
- wave C from right will stop at point -1 m
- wave D from right will stop at point 0 m
Displacement of point P = (amplitude B + amplitude C) + (amplitude A + amplitude D)
Displacement of point P= (2cm - 2cm) + (2 cm - 4cm)= -2cm
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