Answer:
Adding a concentrated salt solution to the solution containing cheek cells will cause the water to move outside the cells by osmosis. The cheek cells are at a higher water potential than the outside solution so, the water will move out, down the water potential gradient, until an equilibrium is reached.
Hope that answers the question, have a great day!
Answer:
If you spill an acid or base on your skin, immediately wash well with water. Strong bases react with the oils in your skin to produce a soapy feeling layer. Rinse until well after that feeling is gone. Do not attempt to neutralize a spill on your skin.
Answer:
A. .2 M/s2
Explanation:
Force = mass x acceleration
The force is given as 20 N and tthe mass as 100kg
Solve for accelartion as 20/100 = .2
(Also wrong subject this is a Physics problem)
Answer:
1.59 x 10⁻²⁵ J.
Explanation:
- The energy of a photon is calculated Planck - Einstein's equation:
E = h ν
, where
E is the energy of the photon,
h is Planck's constant <em>(h = 6.626 x 10
⁻³⁴ J.s)</em>
ν is the frequency of the photon
-
There is a relation between the frequency (ν
) and wave length (λ).
λ.ν = c,
where c is the speed of light in vacuum (c = 3
.0 x 10
⁸ m/s).
λ = 125 cm = 1.25 m.
<em>Now, E = h.c/λ.</em>
∴ E = h.c/λ = (6.626 x 10
⁻³⁴ J.s) (3
.0 x 10
⁸ m/s) / (1.25 m) = 1.59 x 10⁻²⁵ J.
To calculate the <span>δ h, we must balance first the reaction:
NO + 0.5O2 -----> NO2
Then we write all the reactions,
2O3 -----> 3O2 </span><span>δ h = -426 kj eq. (1)
O2 -----> 2O </span><span>δ h = 490 kj eq. (2)
NO + O3 -----> NO2 + O2 </span><span>δ h = -200 kj eq. (3)
We divide eq. (1) by 2, we get
</span>O3 -----> 1.5O2 δ h = -213 kj eq. (4)
Then, we subtract eq. (3) by eq. (4)
NO + O3 -----> NO2 + O2 δ h = -200 kj
- (O3 -----> 1.5 O2 δ h = -213 kj)
NO -----> NO2 - 0.5O2 δ h = 13 kj eq. (5)
eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)
O -----> 0.5O2 <span>δ h = -245 kj eq. (6)
</span>
Add eq. (6) to eq. (5), we get
NO -----> NO2 - 0.5O2 δ h = 13 kj
+ O -----> 0.5O2 δ h = -245 kj
NO + O ----> NO2 δ h = -232 kj
<em>ANSWER:</em> <em>NO + O ----> NO2 δ h = -232 kj</em>
