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3 years ago

During roasting is sulphur dioxide always formed

Chemistry

1 answer:

irakobra [83]3 years ago

6 0

Answer:

In roasting, air in large amounts, sometimes enriched with oxygen, is brought into contact with the sulfide mineral concentrate. This is done at elevated temperatures when oxygen combines with sulfur to form sulfur dioxide and with the metal to form oxides, sulfates and so on. ... Roasting is an exothermic reaction.

calcination as "heating to high temperatures in air or oxygen

Explanation:

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When metallic sodium is dissolved in liquid sodium chloride, electrons are released into the liquid. These dissolved electrons a

qaws [65]

Answer:

The edge of the length is $\mathbf{L = 8.54 \times 10^{-10} \ m}$

Explanation:

From the given information:

The associated energy for a particle in three - dimensional box can be expressed as:

$E_n = \dfrac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2)$

here;

h = planck's constant = $6.626 \times 10^{-34} \ Js$

$n_i$ = the quantum no in a specified direction

m = mass (of particle)

L = length of the box

At the ground state $n_x = n_y = n_z=1$

The energy at the ground state can be calculated by using the formula:

$E_1 =\dfrac{3h^2}{8mL^2}$

At first excited energy level, one of the quantum values will be 2 and the others will be 1.

Thus, the first excited energy will be: 2,1,1

∴

$E_2 =\dfrac{(2^2+1^2+1^2)h^2}{8mL^2}$

$E_2 =\dfrac{(4+1+1)h^2}{8mL^2}$

$E_2 =\dfrac{(6)h^2}{8mL^2}$

The transition energy needed to move from the ground to the excited state is:

$\Delta E= E_2 - E_1$

$\Delta E= \dfrac{6h^2}{8mL^2}- \dfrac{3h^2}{8mL^2}$

$\Delta E= \dfrac{3h^2}{8mL^2}}$ ----- (1)

Recall that:

the wavelength identified with the electronic transition is: 800 nm

800 nm = 8.0 × 10⁻⁷ m

However, the energy-related with the electronic transition is:

$\Delta E =\dfrac{hc}{\lambda}$

$\Delta E =\dfrac{6.626 \times 10^{-34} \times 2.99 \times 10^8}{8.0 \times 10^{-7} }$

$\Delta E =2.48 \times 10^{-19} \ J$

Replacing the value of $\Delta E$ in (1); then:

$2.48 \times 10^{-19}= \dfrac{3h^2}{8mL^2}}$

Making the edge length L the subject of the formula; we have:

$L = \sqrt{\dfrac{3h^2}{8m \times2.48 \times 10^{-19}} }$

$L = \sqrt{\dfrac{3\times (6.626 \times 10^{-34})^2}{8(9.1 \times 10^{-31} ) \times2.48 \times 10^{-19}} }$

$\mathbf{L = 8.54 \times 10^{-10} \ m}$

Thus, the edge of the length is $\mathbf{L = 8.54 \times 10^{-10} \ m}$

5 0

3 years ago

What is [h3o ] in a solution of 0.25 m ch3co2h and 0.030 m nach3co2?

Brilliant_brown [7]

Hello!

To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:

$pH=pKa+log( \frac{[A^{-}] }{[HA]} )$

$pH=4,76+log( \frac{0,030M}{0,25M} ) \\ \\ pH=3,84$

Now, we use the definition of pH and clear [H₃O⁺] from there:

$pH=-log[H_3O^{+}]$

$[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M$

So, the [H₃O⁺] concentration is 0,00014 M

Have a nice day!

4 0

3 years ago

Question 6(Multiple Choice Worth 2 points)

Rina8888 [55]

4.2 × 10²² atoms Al × (1 mol Al / 6.022 × 10²³) = moles Al

Last one: fraction 1 mole Al over 6.022 × 10²³ atoms Al

8 0

3 years ago

What is the maximum number of electrons an expanded octet can hold?

kondaur [170]

First shell hold 2. second shell can hold up to 8. so it should be 8.

7 0

3 years ago

Methylcyclopentene reacting with reacting with sulfuric acid and water

dlinn [17]

Answer:

The product is 1-methylcyclopentanol

Explanation:

The acid-catalyzed hydration of alkenes involves the addition of a water molecule to a C=C double bond.

H₂O + C=C ⟶ H-C-C-OH

An H atom adds to one of the C atoms, and an OH group adds to the other

This reaction follows Markovnikov’s rule — the H adds to the C atom that has more hydrogen atoms, and the OH adds to the more substituted carbon.

The steps of the mechanism are:

The aqueous sulfuric acid generates hydronium ions
The nucleophilic π electrons attack an H atom on the hydronium ion, forming a carbocation on the more substituted C atom.
The lone pair electrons on a water molecule attack the carbocation , forming an oxonium ion.
Another water molecule removes the extra proton.

The product is the alcohol with the OH group on the more substituted carbon — 1-methylcyclopentanol.

7 0

4 years ago