Recent content by Skw

Ok thank you ! Sorry my calculation wasn’t clear, I first reasoned as if it would be 37% v/v and I thought the 1% was the good answer, but then I realised it was w/w so that the good answer was 1.18%. As you suggested, a simpler calculation could be the following: 2.7L at 1.18 kg/L = 2.7*1.18...

2.7L in 97.3L corresponds to a dilution of 1/37 v/v (volume/volume) To prepare a 1% w/w solution : We need 1/37 kg = 0.027 kg so with a density of 1.18 kg/L: 0.027/1.18=2.3% v/v For a 1% w/w HCl solution we need to dilute the stock 2.3% v/v. However, it was diluted 2.7%. The final solution is...

https://www.laboratorynotes.com/molarity-of-37-percent-hydrochloric-acid-hcl/ In this link it defines what is a 37% HCl w/w

No definition

w/w means by weight/weight

1% or 1.18 % (w/w) ?

Hello @BvU, Hello @Mark44, Thank you for your replies! Ah ok thank you for the tip for the concentration in moles per cubic meter, I'll have a thought about it but it looks complicated. Yes I was rounding to the third decimal (for simplicity) at intermediate stages, otherwise, a closer...

I calculate 4.8x10(^10) particles /kg of sand in the sample. Do you find the same ? Is my solution correct ? How many particles do you find ? Thanks in advance !