6g Tris HCl + 100ml dH2O, pH 6.8
to prepare 100ml of 100mM Trissolution: Mol wt of Tris=121.14121.14g in 1000ml ----> 1M12.11g in 100ml -------->1M1M=1000mM121.1g---->1000mM12.11g ----------->100mM1.211g in 100ml and 100mM Tris
No. The heat of reaction for 50mL of each will be multiplied by 2 for 100mL of each since heat of reaction is really on a per mole product basis, and there will be twice as many moles of both HCl and NaOH in 100mL as in 50mL.
1M contains 43.1ml in 500ml, 2M contains 86.2ml in 500ml
10.82
6g Tris HCl + 100ml dH2O, pH 6.8
to prepare 100ml of 100mM Trissolution: Mol wt of Tris=121.14121.14g in 1000ml ----> 1M12.11g in 100ml -------->1M1M=1000mM121.1g---->1000mM12.11g ----------->100mM1.211g in 100ml and 100mM Tris
Preparing 1N HCl for 1L. 1N=1M in HCl. Conc. HCl= 12M M1V1=M2V2 12*V1=1*1000 V1=1000/12 V1=83.33ml 1N HCl= 83.33ml of Conc. HCl in 1L of water 2N HCl= 167ml of Conc. HCl in 1L of water.
No. The heat of reaction for 50mL of each will be multiplied by 2 for 100mL of each since heat of reaction is really on a per mole product basis, and there will be twice as many moles of both HCl and NaOH in 100mL as in 50mL.
1M contains 43.1ml in 500ml, 2M contains 86.2ml in 500ml
10.82
To prepare a 3% acetic (ethanoic) acid solution, you must first standardise the ethanoic acid. This question assumes that has already been done. To make the concentration 3%, there must be 30g per 1000mL (30gL-1). Dissolve or mix in 30g of ethanoic acid per 1000ml (1L). Pipette this into aliquots of 100mL samples. You now have a 3% ethanoic/acetic acid solution.
Dilution from stock solution: Ndilution*Vdilution = Nstock*Vstock Vstock = (0.5N*100ml) / 5.0N = 10ml so take 10ml HCl from 5.0N stock solution + 90ml H2O
Prepare HCl 1 M by HCl concentration 37 % HCl concentration 37 % have density =1.19 g/ml HCl 1 M use HCl 37 % 82.81 ml make volume with water to 1 liter
7.3gm
5g TCA per 100ml of water
You drop conc. HCl into conc. H2SO4 - this releases HCl gas which you can bubble through dry ether... Or, you use an HCl cylinder.