Answer: Air
Hope this helps
Answer:
B = basic
Explanation:
Given data:
[OH⁻] = 5.35×10⁻⁴M
pH = ?
Solution:
pOH = -log[OH⁻]
pOH = - [5.35×10⁻⁴]
pOH = 3.272
it is known that,
pH + pOH = 14
pH = 14- pOH
pH = 14 - 3.272
pH = 10.728
The acidic pH is range from zero to less than 7 while 7 pH is neutral and above 7 the pH is basic. So, the given solution is basic.
When the reaction equation is:
CaSO3(s) → CaO(s) + SO2(g)
we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.
to get the moles of SO2 we are going to use the ideal gas equation:
PV = nRT
when P is the pressure = 1.1 atm
and V is the volume = 14.5 L
n is the moles' number (which we need to calculate)
R ideal gas constant = 0.0821
and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K
so, by substitution:
1.1 * 14.5 L = n * 0.0821 * 285.5
∴ n = 1.1 * 14.5 / (0.0821*285.5)
= 0.68 moles SO2
∴ moles CaSO3 = 0.68 moles
so we can easily get the mass of CaSO3:
when mass = moles * molar mass
and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol
∴ mass = 0.68 moles* 120 g/mol = 81.6 g
7.86 is the pOH of water at this temperature of 100 degrees celsius.
Option E is the right answer.
Explanation:
Data given:
Kw = 51.3 x 
pOH = ?
we know that pure water is neutral and will have pH pf 7.
The equation for relation between Kw and H+ and OH- ion is given by:
Kw = [H+] [OH-}
here the concentration of H+ ion and OH- ion is equal
so, [H+]= [OH-]
Putting the values in the equation of Kw
pKw = -log[Kw]
pKw = -log [51.3 x
]
pKw = 12.28
since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14
Now, pOH is calculated by using the equation:
14 = pOH + pH
14- 6.14 = pOH
pOH = 7.86
Answer:
There are lots of methods.
Explanation:
Usually, animals like pandas live a shorter lifespan in the wild than in captivity. A little fact, there is only one brown panda in the entire world, so it would be very, very rare to see one. The Smithsonian National Zoo, for example, are working to protect pandas, as well as other species.