3.8mL of 0.42 phosphoric acid is required.
Reaction
2H3PO4 + 3CaCL3 → Ca3(PO)4 + 6HCl
moles CaCl2 =0.16 mol/L x0.010 L = 0.0016 mol
moles of H3Po4
= 0.0016mol of CaCl2 x 2 mole of H3PO4/3mole of CaCl2
= 0.00106 mol
V of H3PO4 = 0.0016/0.42 = 0.0038L = 3.8mL
V of H3PO4=3.8mL
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Answer:
if i remember correctly i beleive its A 1.8 x 10^24
but im not for sure also i think you forgot the 24
Explanation:
Answer:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
![Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D%7D%7B%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D%7D)
.
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both ![[\mathrm{SO_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D)
and ![[\mathrm{O_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D)
will increase if the pressure is increased through compression. However, because 
and 
have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient 
.
As a result, the increase in pressure will have no impact on the value of 
. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.
Answer: heat
Other indicators: color change, odor, gas, temperature change.
