At a temperature of 298 K, the Henry's law constant is 0.00130 M/atm for oxygen. The solubility of oxygen in water 1.00 atm would be calculated as follows:
<span>S = (H) (Pgas) = 0.00130 M / atm x 0.21 atm = 0.000273 M
</span>
At 0.890 atm,
<span>S = (H)(Pgas) = 0.00130 M / atm x 0.1869 atm = 0.00024297 M</span>
<span>
If atmospheric pressure would suddenly change from 1.00 atm to 0.890 atm at the same temperature, the amount of oxygen that will be released from 3.30 L of water in an unsealed container would be as follows</span>
<span>
3.30 L x (0.000273 mol / L) = 0.0012012 mol</span>
3.30 L x (0.00024297 mol / L) = 0.001069068 mol
0.0012012 mol - 0.001069068 mol = 0.000132 mol
Current, I got it right on my quiz
<u>Answer:</u>
At time 2t the paint ball is at 8 cm to the right and 16 cm to the bottom
<u>Explanation:</u>
We have equation of motion ,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
Considering the horizontal motion of paint ball
Distance traveled during time t = 4 cm
Initial velocity = u m/s
Acceleration = 0 
So 
Now at time 2t,

So horizontal distance traveled in time 2t = 8 cm to the right
Now considering the vertical motion of paint ball
Distance traveled during time t = 4 cm
Initial velocity = 0 m/s
Acceleration = -g 

At time 2t,

So vertical distance traveled in time 2t = 16 cm to the bottom
Answer:
Angle between incident ray and reflected ray will be 104°
Explanation:
We have given angle of incidence = 52 °
From law of reflection angle of incidence will be equal to angle of reflection
So angle of reflection will be also 52°
We have to find the angle between incident ray and reflected ray
As the incidence angle and reflected angle both is from normal of the surface and opposite to each other
So angle between incident ray and reflected ray will be 52°+52° = 104°
Answer:
Explanation:
Acceleration is the time rate of change of velocity.
Acceleration and velocity are vectors
If east and north are the positive directions, the east moving vector is reduced to zero and the north moving vector increases from zero to 4 m/s.
There are 3 hours or 10800 seconds between 10 AM and 1 PM
a1 = √((-4)² + 4²) / 10800 = (√32) / 10800 m/s² ≈ 4.2 x 10⁻⁴ m/s²
There are 14400 seconds between 10 AM and 2 PM
The velocity changes are still the same
a2 = √((-4)² + 4²) / 10800 = (√32) / 14400 m/s² ≈ 3.9 x 10⁻⁴ m/s²