Estimate the number of photons emitted per second from 1.0 cm2 of a person's skin if a typical emitted photon has a wavelength o
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The density of 53.4 wt aqueous NaOH solution is 0.809 g/ml
Given data:
- The mass percent of NaOH is 53.4.
- Volume of NaOH diluted is 16.7 ml.
- The volume of diluted solution is 2.00 L =2000 ml.
- Concentration of diluted solution is 0.169 M.
First, we find the initial concentration of NaOH by using the following formulae,
M₁V₁ = M₂V₂
Where,
M₁ is the initial molarity of NaOH
M₂ is the molarity after dilution
V₁ is the initial volume of NaOH
V₂ is the final volume after dilution.
Substituting the values,
M₁ × 16.7 ml = 0.169 M × 2000 ml,
M₁ =
M₁ = 20.2 M.
Thus, the initial concentration of NaOH is 20.2 M.
we know, 1 M solution contains 1 mol of substance present in 1 L solution,
Thus, 20.2 M solution will have 20.2 mols of NaOH.
Now, we can find the mass of NaOH by using the number of moles and molar mass.
- molar mass of NaOH is 40 g/mol.
Mass = no. of moles × molar mass
= 20.2 mol × 40 g/mol
= 808 g.
Thus, the mass of NaOH is 808g.
53.4 wt of NaOH means 53.4 g of NaOH in a 100 g solution,
Thus, 808 g of NaOH will be present in ,
⇒
⇒ 1513.1 g
Now, Convert the grams of NaOH to milliliters, using the density of NaOH at room temperature.
- The density of NaOH at room temperature is 1.515 g/ml,
Density =
⇒ 1.515 g/mol =
⇒ volume =
⇒ volume = 998.7 ml.
Thus, the volume of NaOH is 998.7 ml.
Hence, we know,
- the mass of NaOH is 808 g
- the volume of NaOH is 998.7 ml
Substituting the values,
Density = 808 g / 998.7 g/ml
⇒ Density = 0.809 g/ml
Thus, the density of 53.4 wt aqueous NaOH is 0.809 g/ml.
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Answer:
The magnitude of the electric field inside the membrane is 8.8×10⁶V/m
Explanation:
The electric field due to electric potential at a distance Δs is given by
E=ΔV/Δs
We have to find the magnitude electric field in the membrane
Ecell= -ΔV/Δs
The magnitude of the electric field inside the membrane is 8.8×10⁶V/m
Answer:
Explanation:
As we know that truck is moving up the hill
So here we have two forces on the crate placed on the truck
1) gravitational force along the floor
2) Friction force along the floor
Now here if the truck is moving with maximum possible acceleration so that there is no slipping of the crate
So here we will have