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2 years ago

How are exothermic and endothermic reactions linked in the process of refining metal ore?

1 answer:

6 0

The intended sense is that of a reaction that depends on absorbing heat if it is to proceed. The opposite of an endothermic process is an exothermic process, one that releases "gives out" energy in the form of heat

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HURRY I WILL GIVE YOU BRAINLIEST

Amiraneli [1.4K]

dim? im not so fluent in this but i did research yesterday

6 0

2 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Someone help me pls will make you as brain

jonny [76]

Answer:

i, ii, and iii

Explanation:

3 0

2 years ago

Read 2 more answers

Question 3 (1 point) The extinction coefficient for copper sulfate in aqueous solution is 12 M-1.cm-1 at 800 nm. If the absorban

Troyanec [42]

Answer:

The concentration of the copper sulfate solution is 83 mM.

Explanation:

The absorbance of a copper sulfate solution can be calculated using Beer-Lambert Law:

A = ε . c . <em>l</em>

where

ε is the extinction coefficient of copper sulfate (ε = 12 M⁻¹.cm⁻¹)

c is its molar concentration (what we are looking for)

l is the pathlength (0.50 cm)

We can use this expression to find the molarity of this solution:

$c=\frac{A}{\epsilon.l } =\frac{0.5}{12M^{-1}cm^{-1}0.50cm } =0.083M=83mM$

3 0

2 years ago

In which conditions combination and displacement reaction takes place explain with example.

expeople1 [14]

<h2>Answer:</h2>

In <u>Combination reaction</u>, two or more elements combined to form one compound of different properties.

C(s) + O2(g) ⇢ CO2(g).
H2(g) + O2(g) ⇢ H20(l).

In <u>Displacement reation</u>, the high reactive element displaces the low reactive element and formed compound of different properties.

Fe(s) + CuSo4(aq) ⇢ FeSo4(aq) + Cu(s).
AgNO3(aq) + Cu(s) ⇢ CuNO3(aq) + Ag(s).

8 0

2 years ago

Read 2 more answers