P2 = 54.6 kPa
Explanation:
Given:
V1 = 10.0 L. V2 = 50.0 L
P1 = 273 kPa. P2 = ?
We can use Boyle's law to solve this problem.
P1V1 = P2V2
Solving for P2,
P2 = (V1/V2)P1
= (10.0 L/50.0 L)(273 kPa)
= 54.6 kPa
Missing question:
Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that:
1. The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose.
2. The temperature of the air is constant (body temperature).
3. The air acts as an ideal gas.
4. Salt water has an average density of around 1.03 g/cm^3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm.
T = 37°C = 310 K.
p₁ = 2,5 atm = 253,313 kPa.
p₂ = 1 atm = 101,325 kPa.
Ideal gas law: p·V = n·R·T.
n₁ = 253,313 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₁ = 0,589 mol.
n₂ = 101,325 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₂ = 0,2356 mol.
Δn = 0,589 mol - 0,2356 mol = 0,3534 mol.
<span>Kw = water autoionization constant=1.0 x 10-14 @ 25 °C
Kw = [H3O+] [OHÂŻ]
1.0 x 10-14= [x][x]
x=1.00 x 10ÂŻ7 M</span>
Answer:
K2 +Br ->2KBr
K + I ->KI
actually I don't know the e option but I had tried can u pls balance it urself
Answer:
D
Explanation:
The correct answer that gives a good example of how the same chemical used to help people can harm the environment would be when <u>excess fertilizers run off into rivers and create a dead zone.</u>
<em>Fertilizers are usually applied to crops as an agricultural practice by farmers to enhance the yield of the crops and boost food or raw materials production. However, excess fertilizer gets washed into nearby water bodies and ends up creating a low-oxygen zone that is incapable of supporting lives within the water body - also known as </em><em>dead zones.</em>
The correct option is, therefore, D.
