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Vikentia [17]
3 years ago
14

Suppose you are titrating an acid solution with a base solution of known concentration. To calculate the concentration of the ac

id solution, use three steps.
1. Use the -------------- { (a) delivered volume (b) volume reading} of base --------------- { (a) after (b) to reach (c) before} the endpoint and the known concentration of the base solution to find the ------------ { (a) moles (b) mass (c) concentration} of base used.
2. Use the ------------ { (a) molar mass (b) dilution equation (c) molar ratio} to find the moles of acid from the moles of base.
3. Divide the moles of acid by the volume of -------------- { (a) base solution before (b) acid solution before (c) acid solution after} the titration to find the concentration of acid.
Chemistry
1 answer:
Arturiano [62]3 years ago
4 0

Answer:

i would say use 2

Explanation:

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The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon a
DIA [1.3K]

Answer:

Empirical formula = C5H4

Molecular formula = C10H8

Explanation:

When the 3000 mg of naphthalene are burned they produce 10.3 mg of CO2. Knowing the unbalanced equation of the combustion of naphthalene, we have:

CxHy + O2 = CO2 + H2O

We calculate the molar composition of the sample. We look for the molecular weights in the periodic table:

CO2 = 12,011 + 2 (15,999) = 44,009 g

Mol C = 10.3 mg * (1 mol CO2 / 44.009 g CO2) * (1 mol C / 1 mol CO2) = 0.234 mmol C

Mass C = 0.234 mmol C * (12.011 g C / 1 mol C) = 2.8105 mg C

Mass H = 3 mg - 2.8105 mg = 0.1895 mg H

Mol H = 0.1895 mg H * (1 mol H / 1,008 g H) = 0.188 mmol H

To calculate the empirical formula, we must divide the number of moles of each element by the smallest number of moles, in this case, of hydrogen:

C = 0.2340 mmol C / 0.1895 mol H = 1.25

H = 0.1895 mmol H / 0.1895 mmol H = 1

We multiply the coefficients by 4, and we have the empirical formula:

C1.25 * 4H1 * 4 = C5H4

The molecular formula is equal to (C5H4)m, where m is calculated by the molecular and empirical mass ratio, as follows:

Empirical mass = (5 * 12.011) + (4 * 1.008) = 64.09 g

m = 130 g / 64.09 g = 2.02 = 2

Therefore we have the molecular formula:

(C5H4)2 = C10H8

4 0
3 years ago
PLEASE ASNWR ASAP
Vikki [24]
34.95 atm

lol i hope i’m not too late



4 0
3 years ago
The reaction 2KBr(s) →+ Br2(l) + 2K(s) is a <br>Decomposition reaction
klasskru [66]
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5 0
3 years ago
How many meters does daniel have to walker
Tcecarenko [31]

14,200 because all you have to do to solve this is multiply 14.2 kilometers by 1,000 meters to find the distance that he walks.

8 0
4 years ago
Read 2 more answers
Octane has a density of 0.703 g/ml. Calculate the mass of CO2(g) produced by burning one
STatiana [176]

The mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

The given parameters:

  • <em>Density of the octane, ρ = 0.703 g/ml</em>
  • <em>Volume of octane, v = 3.79 liters</em>

<em />

The mass of the octane burnt is calculated as follows;

m = \rho V\\\\m = 0.703 \ \frac{g}{ml} \times 3.79 \ L \ \frac{1000 \ ml}{L} \\\\m = 2,664.37 \ g

The combustion reaction of octane is given as;

2C_8H_{18} +  \ 25O_2 \ --> \ 16CO_2 \ + \ 18H_2O

From the reaction above:

228.46 g of octane -------------------> 704 g of  CO₂ gas

2,664.37 of octane --------------------> ? of CO₂ gas

= \frac{2,664.37 \times 704}{228.46} \\\\= 8,210.3 \ g\\\\= 8.21 \ kg

Thus, the mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

Learn more about combustion of organic compounds here: brainly.com/question/13272422

8 0
3 years ago
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