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soldi70 [24.7K]
3 years ago
7

After substantial heating, 6.25 g of iron produced 18.00 g of a compound with chlorine. The empirical formula is:

Chemistry
1 answer:
babunello [35]3 years ago
3 0

Answer:

Option A. FeCl3

Explanation:

The following data were obtained from the question:

Mass of iron (Fe) = 6.25g

Mass of the compound formed = 18g

From the question, we were told that the compound formed contains chlorine. Therefore the mass of chlorine is obtained as follow

Mass of chlorine (Cl) = Mass of compound formed – Mass of iron.

Mass of chlorine (Cl) = 18 – 6.25

Mass of chlorine (Cl) = 11.75g

The compound therefore contains:

Iron (Fe) = 6.25g

Chlorine (Cl) = 11.75g

The empirical formula for the compound can be obtained by doing the following:

Step 1:

Divide by their molar mass

Fe = 6.25/56 = 0.112

Cl = 11.75/35.5 = 0.331

Step 2:

Divide by the smallest

Fe = 0.112/0.112 = 1

Cl = 0.331/0.112 = 3

The empirical formula for the compound is FeCl3

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It’s A non of these

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5 0
2 years ago
What is the limiting reactant if 0.5 g Al is reacted with 3.5 g CuCl2?
DIA [1.3K]
The balanced equation for the above reaction is 
2Al + 3CuCl₂ --> 2AlCl₃ + 3Cu
stoichiometry of Al to CuCl₂ is 2:3
limiting reactant is when the reactant is fully consumed in the reaction therefore amount of product formed depends on amount of limiting reactant present. 
number of Al moles  - 0.5 g  / 27 g/mol = 0.019 mol
number of CuCl₂ moles - 3.5 g / 134.5 g/mol = 0.026 mol
if Al is the limiting reactant 
if 2 mol of Al reacts with 3 mol of CuCl₂
then 0.019 mol of Al reacts with - 3/2 x 0.019 = 0.029 mol of CuCl₂
but only 0.026 mol of CuCl₂ is present 
therefore CuCl₂ is the limiting reactant 
and 0.026 mol of CuCl₂ reacts with - 0.026/3 x 2 = 0.017 mol of Al is required
but 0.019 mol of Al is present 
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6 0
3 years ago
4) The initial rate of the reaction between substances P and Q was measured in a series of
ASHA 777 [7]

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

rate = k[P]^{2} [Q]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

k=\frac{rate}{[P]^{2}[Q] } \\  =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4

Second row:

2. Rate value:

rate =0.4* (0.10)^{2} * (0.10)\\\\        =4.0*10^-3mol.dm^-3.s^-1

3.Third row:

[Q]=\frac{rate}{k.[P]^{2} } \\     =9.6*10^-3 / (0.4 *(0.40)^{2} \\    =0.15mol.dm^{-3}

4. Fourth row:

[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}

6 0
2 years ago
The nonvolatile, nonelectrolyte testosterone, C19H28O2 (288.40 g/mol), is soluble in benzene C6H6. Calculate the osmotic pressur
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Answer:

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T is temperature of benzene solution = 298 K

MW is molecular weight of testosterone = 288.40 g/mol

V is volume of benzene solution = 286 ml = 286 cm^3

P = 12.9×82.057×298/288.4×286 = 3.824 atm

6 0
3 years ago
Which type of equipment would be used to precisely measure 26.0 mL of dilute hydrochloric acid?
VikaD [51]
The type of equipment that would be used to precisely measure 26.0 mL of dilute hydrochloric acid would be C. 50 mL graduated cylinder.
D doesn't have enough mLs to measure this, and A and B have too many. 
8 0
3 years ago
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