When [HClO]= 0.01 M & [ClO]^- = 0.03M & NaOH = 0.003 mol and
Ka = 3.5 x 10^-8
and the equation is:
HClO + NaOH → Na^+ + ClO^-
initial 0.01 0.003 0.03 0.03
-0.003 - 0.003 +0.003 + 0.003
Final 0.007 0 0.033 0.033
We can get the PH from this formula:
PH = Pka + ㏒[conjugent base/weak acid]
PH = -㏒Ka + ㏒[Na]/[HClO]
= - ㏒ 3.5x10^-8 + ㏒(0.033/0.007)
= 8.13
Answer:
95.54mL
Explanation:
To calculate the volume that would have that mass, what is needed to be done is to use the formula that relates mass, density and volume.
Mathematically,
mass = density * Volume
But in this particular question we are to calculate the volume, we already have the mass and the density.
Rearranging the equation gives the following:
volume = mass/density
Our mass here is 75.0g while our density is 0.785g/mL
Hence, volume = 75/0.785 = 95.54mL
H20 and c20 were brothers and sisters before ch3ch2nh2 even existed
Answer:
Inflamabilidad y características de la llama: el hidrógeno es inflamable en el aire en un amplio rango de concentraciones y arde, en ausencia de impurezas, con una llama casi invisible. Energía de ignición: el hidrógeno puede entrar en ignición con una cantidad de energía muy pequeña.
Explanation:
espero te sirva✍️
First, calculate the number of moles of sodium present with the given mass,
31.5 g of sodium x (1 mol sodium/ 23 g sodium) = 1.37 mol sodium
It is given in the equation that for every 2mols of sodium, one mol of H2 is produced.
mols of H2 = (1.37 mols sodium)(1 mol H2/ 2 mols sodium)
mols of H2 = 0.685 mols H2
Then, at STP, 1 mol of gas = 22.4 L.
volume of H2 = (0.685 mols H2)(22.4 L / 1 mol)
volume of H2 = 15.34 L
Answer: 15.34 L
