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3 years ago

A student mixed a small amount of iron filings and sulphur powder in a dish. He could not

Chemistry

1 answer:

Temka [501]3 years ago

7 0

Answer:

$\huge \boxed{\mathrm{Carbon \ disulphide}}$

Explanation:

Carbon disulphide is the liquid that can be used to separate iron fillings and sulphur powder.

When carbon disulphide is poured into the dish, the sulphur powder gets easily dissolved in the carbon disulfide. The iron fillings are left to settle on the bottom of the dish.

The iron fillings can get seperated through filtration. When the mixture of sulphur powder and carbon disulphide gets completely evaporated, the sulphur powder is left over.

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A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

3 years ago

The properties of compounds are (what?) to the elements that make it up. *

Lana71 [14]

Answer:A compound is a unique substance that forms when two or more elements combine chemically. A compound always has the same elements in the same proportions. The properties of compounds may be very different from the properties of the elements that form them. Some compounds form rigid frameworks called crystals.

Explanation:

8 0

3 years ago

Read 2 more answers

What mass of iron(II) oxide must be used in the reaction given by the equation below to release 44.7 kJ? 6FeO(s) + O2(g) => 2

zavuch27 [327]

<u>Answer:</u> The mass of iron (II) oxide that must be used in the reaction is 30.37

<u>Explanation:</u>

The given chemical reaction follows:

$6FeO(s)+O_2(g)\rightarrow 2Fe_3O_4(s);\Delta H^o=-635kJ$

By Stoichiometry of the reaction:

When 635 kJ of energy is released, 6 moles of iron (II) oxide is reacted.

So, when 44.7 kJ of energy is released, $\frac{6}{635}\times 44.7=0.423mol$ of iron (II) oxide is reacted.

Now, calculating the mass of iron (II) oxide by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of iron (II) oxide = 0.423 moles

Molar mass of iron (II) oxide = 71.8 g/mol

Putting values in above equation, we get:

$0.423mol=\frac{\text{Mass of FeO}}{71.8g/mol}\\\\\text{Mass of FeO}=(0.423mol\times 71.8g/mol)=30.37g$

Hence, the mass of iron (II) oxide that must be used in the reaction is 30.37

7 0

3 years ago

Using your knowledge of reactions and mechanisms, draw the mechanism for the formation of 2-methylpropene from 2-methyl-2-propan

BARSIC [14]

Answer:

Here's what I get

Explanation:

The reaction is an E1 elimination of an alcohol to form an alkene. It has three steps:

1. Protonation

The alcohol is protonated with aqueous sulfuric acid to convert it into a better leaving group.

2. Loss of the leaving group

A water molecule leaves in a unimolecular process to form a stable 3° carbocation.

3. Loss of an α-hydrogen

A water molecule removes an α-hydrogen, forming 2-methylpropene and regenerating the original hydronium ion.

4 0

3 years ago

What is the molar mass of BaBr2? A. 217.2 g/mol B. 297.1 g/mol C. 354.5 g/mol D. 434.4 g/mol

qaws [65]

Answer: Option (B) is the correct answer.

Explanation:

Molar mass is defined as the sum of masses of all the atoms present in a compound.

For example, atomic mass of barium is 137.32 g/mol and atomic mass of bromine is 79.90 g/mol.

Therefore, molar mass of $BaBr_{2}$ will be as follows.

Molar mass = atomic mass of Ba + $2 \times atomic mass of Br$

= 137.32 g/mol + $2 \times 79.90 g/mol$

= 297.12 g/mol

Hence, we can conclude that molar mass of [tex]BaBr_{2}[tex] is 297.12 g/mol.

8 0

3 years ago

Read 2 more answers