Answer:
2.59g
Explanation:
Using Avagadro's law equation as follows:
V1/n1 = V2/n2
Where;
V1 = initial volume (litres)
V2 = final volume (litres)
n1 = first amount of gas in grams
n2 = second amount of gas in grams
According to the information provided in this question, v1 = 93.2L, v2 = 10.4L, n1 = 23.2 g, n2 = ?
Using V1/n1 = V2/n2
93.2/23.2 = 10.4/n2
Cross multiply
93.2 × n2 = 23.2 × 10.4
93.2n2 = 241.28
n2 = 241.28/93.2
n2 = 2.588
n2 = 2.59g
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2M was neutralized by 0.01 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 0.125 mL
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl (Stomach acid)
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 0.125 mL
Answer:
(i) 5-day BOD of the waste is 120 mg/l.
(ii) The ultimate carbonaceous BOD (Lo) is 20 mg/l.
Explanation:
The dilution factor D is 0.05.
The initial DO is 8.0 mg/L and the DO after 5 days is 2.0 mg/L.
The BOD of the waste for an unseeded mixture is
The ultimate carbonaceous BOD (Lo) can be calculated as
Answer:
40 Grams
Explanation:
Fe2O3 is 2 parts Iron 3 parts Oxygen.
Assuming the molecular weights of both are equivalent (<em>ideal elemental law</em>), it is a 2:3 ratio, or 2/5 of the total.
2/5 of 100 is 40, so the answer is
40 grams
Answer:
1.04 moles of AlCl₃
Explanation:
Balanced Equation: 2Al°(s) + 3Cl₂(g) => 2AlCl₃(s)
Approach => convert given data (28 g Al°) to moles. The number of moles AlCl₃ would equal the number of moles aluminum because coefficients for Aluminum and Aluminum Chloride are equal.
moles Al° = (28g)/(27g/mole ) = 1.04 mole Al°
From balanced equation moles AlCl₃ produced equals moles of Al° produced = 1.04 mole AlCl₃ because coefficients of Al° & AlCl₃.