Answer: The molarity of an HCl solution is 0.218 M if 43.6 mL of a 0.125 M NaOH solution are needed to titrate a 25.0 mL sample of the acid.
Explanation:
Given: 
= 43.6 mL, 
= 0.125 M
= 25.0 mL, 
= ?
Formula used to calculate the concentration of acid is as follows.
Substitute the values into above formula.
Thus, we can conclude that the molarity of an HCl solution is 0.218 M if 43.6 mL of a 0.125 M NaOH solution are needed to titrate a 25.0 mL sample of the acid.
Electron structure of sodium:
₁₁Na: 1s²2s²2p⁶3s¹
464 g radioisotope was present when the sample was put in storage
<h3>Further explanation</h3>
Given
Sample waste of Co-60 = 14.5 g
26.5 years in storage
Required
Initial sample
Solution
General formulas used in decay:
t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Half-life of Co-60 = 5.3 years
Input the value :
Mass is equal to moles x molar mass, and the molar mass of C6H12 is 84, therefore the mass is 436.8 g, but 437 rounded to correct significant figures
<h3>Answer:</h3><h3>1865.5g</h3><h3>Explanation:</h3><h3 /><h2> first the chemical formular for ammonium hydroxide is NH4OH</h2><h3>its molarmass is given as N=14H=1O=16 </h3><h3> so we have 14 +1(2) +16+1 =35</h3><h2>also no of moles = mass / molarmass</h2><h3> we have 5.33×10 = mass/35 </h3><h2>therefore mass = 35 ×5.33×10 = 1865.5g</h2>
