Answer:
It contains 0.105 mole cu
Explanation:
Emperic formula is SO subscript 2. Molecular formula would be SO subscript 2 multiply 1 so it's the same answer.
We can set up an ICE table for the reaction:
HClO H+ ClO-
Initial 0.0375 0 0
Change -x +x +x
Equilibrium 0.0375-x x x
We calculate [H+] from Ka:
Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)
Approximating that x is negligible compared to 0.0375 simplifies the equation to
3.0x10^-8 = (x)(x)/0.0375
3.0x10^-8 = x2/0.0375
x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9
x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.
We can now calculate pH:
pH = -log [H+] = - log (3.35 x 10^-5) = 4.47
Answer:
1.209g of MgO participates
Explanation:
In this problem, we have 0.030 moles of MgO that participates in a particular reaction.
And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.
To convert moles to grams we need to use molar mass of the compound:
<em>1 atom of Mg has a molar mass of 24.3g/mol</em>
<em>1 atom of O has a molar mass of 16g/mol</em>
<em />
That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol
And mass of 0.030 moles of MgO is:
0.030 moles MgO * (40.3g/mol) =
<h3>1.209g of MgO participates</h3>
Answer:
mass P4 = 35.998 g
Explanation:
∴ STP: P = 1 atm; T = 298 K
∴ V O2= 35.5 L
⇒ nO2 = P.V / R.T
∴ R = 0.082 atm.L/K.mol
⇒ nO2 = ((1 atm)×(35.5L))/((0.082 atm.L/K.mol)(298K))
⇒ nO2 = 1.453 mol O2
⇒ mol P4 = (1.453 molO2)×(mol P4/ 5molO2) = 0.2906 mol P4
∴ Mw P4 = 123.895 g/mol
⇒ mass P4 = (0.2906 mol P4)×(123.895 g/mol) = 35.998 g P4
