to neutralize 1 mole of H2 S o4 we need one mole of any if we are having 50 grams of H2 S o4 it means the mole of H2 S o4 in 50 gram will be 50×40)/98 hence
utilising 50 grams of H2 S o4 we need approximately 20. 5 gm of Naoh
The concentration of lead nitrate is 3.48 M.
<u>Explanation:</u>
The molarity can be found by dividing moles of sucrose by its volume in litres. We can find the number of moles of sucrose by dividing the given mass by its molar mass. Now we can find the moles as,
Here mass of Pb(NO₃)₂ is 380 g
Molar mass of Pb(NO₃)₂ is 331.2 g/mol
Number of moles = 
= 
= 1.15 moles
Volume in Litres = 330 ml = 0.33 L
Molarity = 
= 3.48 mol/L or 3.48 M
So the concentration of lead nitrate is 3.48 M.
The volume of the gas at STP = 35.01 L
<h3>Further explanation</h3>
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure).
In general, the gas equation can be written

where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
V=17.4 L
T = 23 + 273 = 296 K
P = 2.18 atm

The volume of the gas occupy at STP :

0,5 kilometers = 500 metres
500<600
Maria ran farther
To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.
.675 M x V1 = .25 M x 1.3 L
V1 = 0.48 L or 480 mL