Answer:
104.84 moles
Explanation:
Given data:
Moles of Boron produced = ?
Mass of B₂O₃ = 3650 g
Solution:
Chemical equation:
6K + B₂O₃ → 3K₂O + 2B
Number of moles of B₂O₃:
Number of moles = mass/ molar mass
Number of moles = 3650 g/ 69.63 g/mol
Number of moles = 52.42 mol
Now we will compare the moles of B₂O₃ with B from balance chemical equation:
B₂O₃ : B
1 : 2
52.42 : 2×52.42 = 104.84
Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.
Answer:
A. The moles of H(aq) equal the moles of OH
Explanation:
Thats what my chemistry teacher said Just trying to help out since theres no other answers.
Answer:
B) irreversible process
Explanation:
The process given here is irreversible.
Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
Answer:
178 grams
Explanation:
<em>It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³).</em>
<em><u>Using cross multiplication:</u></em>
1.0 mol contains → 6.022 x 10²³ molecules.
??? mol contains → 6.3 x 10²⁴ molecules.
∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(6.3 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 10.46 mol.
<em>∴ The no. of grams of NH₃ present = no. of moles x molar mass</em> = (10.46 mol)(17.0 g/mol) = <em>177.8 g ≅ 178 g.</em>
