Answer:
The mass of 2,50 moles of NaCl is 146, 25 g.
Explanation:
First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate the mass of 2.50 moles of compound, making a simple rule of three:
Weight NaCl= Weight Na + Weight Cl= 23 g+ 35,5 g= 58, 5 g/ mol
1 mol ------ 58, 5 g
2,5 mol---x= (2,5 mol x 58, 5 g)/ 1 mol = <u>146, 25 g</u>
Here is the complete question.
Benzalkonium Chloride Solution ------------> 250ml
Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.
Sig: Dilute 10ml to a liter and apply to affected area twice daily
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL
Answer:
(B) 29.4 mL
Explanation:
1 L = 1000 mL
1:200 solution implies the
in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be 
200mL × 1g = 1000 mL × x(g)
x(g) = 
x(g) = 0.2 g
That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.
∴ 
y(g) = 
y(g) = 5g of benzalkonium chloride.
Now, at 17%
concentrate contains 17g/100ml:
∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;
= 
z(mL) = 
z(mL) = 29.41176 mL
≅ 29.4 mL
Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride
The particles that make up the atomic nucleus of all atoms are both protons and neutrons.
Answer:
The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 
Explanation:
Given :
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

= Rydberg energy
n = principal quantum number of the orbital
Energy of 11th orbit = 

Energy of 10th orbit = 

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.


(Planck's' equation)


The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 