Reactivity trends of halogen:
1) Melting point and boiling points increased down the
group
2) Colour becomes darker.
E.g. Fluorine (pale yellow)
Chlorine (yellowish-green)
Bromine (reddish-brown)
Iodine (purplish-black)
Astatine (black)
3) The reactivity decreases down the group.
Reactivity:
F > Cl > Br > I > At
Answer:
water, when the metastable state is reached, is cooled below the zero temperature. It freezes abruptly. this is called metastable. They are not at equilibrium per se; as at negative temperatures the only equilibrium state of water is ice.
Explanation:
The equation structure for the above mentioned reaction can be written as
<u>Explanation:</u>
Considering the above reaction, When Boron sulfide, reacts with water more violently to form boric acid and hydrogen sulfide gas.
In order to balance the equation, we can do as follows.There are 2 B - atoms on both sides of the equation, but only 2 H - atoms, and one O - atom on LHS, so we have to balance it by putting 6 in front of water and 2 in front of Boric acid and 3 in front of hydrogen sulphide gas, so that we have 2 B - atoms, 3 - S atoms, 12 H - atoms on both sides of the equation, and it is balanced. Balanced equation is given as,
Thus a Balanced equation of the above mentioned reaction is written.
Answer: 4.96 moles
Explanation:
C5H12 is the chemical formula for pentane, the fifth member of the alkane family.
Given that,
number of moles of C5H12 = ?
Mass in grams = 357.4 g
Molar mass of C5H12 = ?
To get the molar mass of C5H12, use the atomic mass of carbon = 12g; and Hydrogen = 1g
i.e C5H12 = (12 x 5) + (1 x 12)
= 60g + 12g
= 72g/mol
Now, apply the formula
Number of moles = Mass / molar mass
Number of moles = 357.4g / 72g/mol
= 4.96 moles
Thus, 4.96 moles of C5H12 that are contained in 357.4 g of the compound.
Answer:
a) 0 J
b) -2.67x10² J
c) -2.09x10³ J
Explanation:
For an isothermic expansion (with constant temperature) the work (W) is :
W = -pΔV, where p is the pressure and ΔV the volume variation. The minus signal is used because the compression is positive and ΔV is negative (Vf < Vi).
a) In vacuum, the relative pressure is 0 atm, so the work:
W = -0x(5.7 - 1.3)
W = 0 J
b) For a constant pressure of 0.60 atm
W = -0.6atmx(5.7 - 1.3)L = -2.64 L.atm
1 L.atm = 101.3 J
W = -2.64x101.3 = -2.67x10² J
c) For a pressure of 4.7 atm
W = -4.7atmx(5.7 - 1.3) L = - 20.68 atm.L
1 atm.L = 101.3 J
W = -20.68x101.3 = -2.09x10³ J
