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4 years ago

What happens when sodium and sulfur combine

Chemistry

1 answer:

Eduardwww [97]4 years ago

7 0

Answer:

Sodium sulfide is the chemical compound with the formula Na2S, or more commonly its hydrate Na2S·9H2O. Both the anhydrous and the hydrated salts are colorless solids. They are water-soluble, giving strongly alkaline solutions. When exposed to moist air, Na2S and its hydrates emit hydrogen sulfide, which smells like rotten eggs. Some commercial samples are specified as Na2S·xH2O, where a weight percentage of Na2S is specified. Commonly available grades have around 60% Na2S by weight, which means that x is around 3. Such technical grades of sodium sulfide have a yellow appearance owing to the presence of polysulfides. These grades of sodium sulfide are marketed as 'sodium sulfide flakes'.

Contents

1 Structure

2 Production

3 Reactions with inorganic reagents

4 Uses

4.1 Reagent in organic chemistry

5 Safety

6 References

Structure

Na2S adopts the antifluorite structure,[2][3] which means that the Na+ centers occupy sites of the fluoride in the CaF2 framework, and the larger S2− occupy the sites for Ca2+.

Production

Industrially Na2S is produced by carbothermic reduction of sodium sulfate often using coal:[4]

Na2SO4 + 2 C → Na2S + 2 CO2

In the laboratory, the salt can be prepared by reduction of sulfur with sodium in anhydrous ammonia, or by sodium in dry THF with a catalytic amount of naphthalene (forming sodium naphthalenide):[5]

2 Na + S → Na2S

Reactions with inorganic reagents

The sulfide ion in sulfide salts such as sodium sulfide can incorporate a proton into the salt by protonation:

S2−

+ H+ → SH−

Because of this capture of the proton ( H+), sodium sulfide has basic character. Sodium sulfide is strongly basic, able to absorb two protons. Its conjugate acid is sodium hydrosulfide (SH−

). An aqueous solution contains a significant portion of sulfide ions that are singly protonated.

S2−

+ H

2O {\displaystyle {\ce {<=>>}}}{\displaystyle {\ce {<=>>}}} SH−

+ OH−

(1)

SH−

+ H

2O {\displaystyle {\ce {<<=>}}}{\displaystyle {\ce {<<=>}}} H

2S + OH−

(2)

Sodium sulfide is unstable in the presence of water due to the gradual loss of hydrogen sulfide into the atmosphere.

When heated with oxygen and carbon dioxide, sodium sulfide can oxidize to sodium carbonate and sulfur dioxide:

2 Na2S + 3 O2 + 2 CO

2 → 2 Na2CO3 + 2 SO2

Oxidation with hydrogen peroxide gives sodium sulfate:[6]

Na2S + 4 H2O2 → 4 H

2O + Na2SO4

Upon treatment with sulfur, polysulfides are formed:

2 Na2S + S8 → 2 Na2S5

Uses

Sodium sulfide is primarily used in the kraft process in the pulp and paper industry.

It is used in water treatment as an oxygen scavenger agent and also as a metals precipitant; in chemical photography for toning black and white photographs; in the textile industry as a bleaching agent, for desulfurising and as a dechlorinating agent; and in the leather trade for the sulfitisation of tanning extracts. It is used in chemical manufacturing as a sulfonation and sulfomethylation agent. It is used in the production of rubber chemicals, sulfur dyes and other chemical compounds. It is used in other applications including ore flotation, oil recovery, making dyes, and detergent. It is also used during leather processing, as an unhairing agent in the liming operation.

Reagent in organic chemistry

Alkylation of sodium sulfide give thioethers:

Na2S + 2 RX → R2S + 2 NaX

Even aryl halides participate in this reaction.[7] By a broadly similar process sodium sulfide can react with alkenes in the thiol-ene reaction to give thioethers. Sodium sulfide can be used as nucleophile in Sandmeyer type reactions.[8] Sodium sulfide reduces1,3-dinitrobenzene derivatives to the 3-nitroanilines.[9] Aqueous solution of sodium sulfide can be refluxed with nitro carrying azo dyes dissolved in dioxane and ethanol to selectively reduce the nitro groups to amine; while other reducible groups, e.g. azo group, remain intact.[10] Sulfide has also been employed in photocatalytic applications.[11]

Explanation:there you go

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There are two possible isomers of the product acetal that could conceivably be formed.

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Answer:

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3 years ago

Which of the following is a scientific question about the cuttlefish?

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The answer is A. How does the cuddle fish change its colors ?

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3 years ago

Read 2 more answers

Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant f

umka21 [38]

Answer:

1) 0.00498 mol Cu.

2) 0.00000374 mol CO₂

Explanation:

Question 1)

We want to convert 3.00 * 10²¹ copper atoms into moles. Note that 3.00 is three significant figures.

Recall that by definition, one mole of a substance has exactly 6.022 * 10²³ amount of that substance. In other words, we have the ratio:

$\displaystyle \frac{1\text{ mol}}{6.022\times 10^{23} \text{ Cu}}$

We are given 3.00 * 10²¹ Cu. To cancel out the Cu, we can multiply it by our above ratio with Cu in the denominator. Hence:

$\displaystyle 3.00 \times 10^{21} \text{ Cu} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} \text{ Cu}}$

Cancel like terms:

$=\displaystyle 3\times 10^{21} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} }$

Simplify:

$\displaystyle = \frac{3\text{ mol Cu}}{6.022 \times 10^{2}}$

Use a calculator:

$= 0.004981... \text{ mol Cu}$

Since the resulting answer must have three significant figures:

$= 0.00498\text{ mol Cu}$

So, 3.00 * 10²¹ copper atoms is equivalent to approximately 0.00498 moles of copper.

Question 2)

We want to convert 2.25 * 10¹⁸ molecules of carbon dioxide into moles. Note that 2.25 is three significant digits.

By definition, there will be 6.022 * 10²³ carbon dioxide molecules in one mole of carbon dioxide. Hence:

$\displaystyle \frac{6.022 \times 10^{23} \text{ CO$_2$}}{1\text{ mol CO$_2$}}$

To cancel the carbon dioxide from 2.25 * 10¹⁸, we can multiply it by the above ratio with the carbon dioxide in the denominator. Hence:

$\displaystyle 2.25\times 10^{18} \text{ CO$_2$} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23} \text{ CO$_2$}}$

Cancel like terms:

$\displaystyle= 2.25\times 10^{18} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23}}$

Simplify:

$\displaystyle = \frac{2.25 \text{ mol CO$_2$}}{6.022\times 10^5}}$

Use a calculator:

$=0.000003736...\text{ mol CO$_2$}$

Since the resulting answer must have three significant figures:

$= 0.00000374\text{ mol CO$_2$}$

So, 2.25 * 10¹⁸ molecules of carbon dioxide is equivalent to approximately 0.00000374 moles of carbon dioxide.

4 0

3 years ago

Read 2 more answers

Need help !!!!! ASAP

Nat2105 [25]

<h2>Hello!</h2>

The answer is:

The temperature will be the same, 37°C.

Since from the statemet we know the first temperature, pressure and volumen of a gas, and we need to calculate the new temperature after the pressure and the volume changed, we need to use the Combined Gas Law.

The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using Boyle's Law, Gay-Lussac's Law and Charles's Law.

The law establishes the following equation:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}$

Where,

$P_{1}$ is the first pressure.

$V_{1}$ is the first volume.

$T_{1}$ is the first temperature.

$P_{2}$ is the second pressure.

$V_{2}$ is the second volume.

$T_{2}$ is the second temperature.

Then, we are given the following information:

$V_{1}=200mL\\P_{1}=4atm\\T_{1}=37\°C\\V_{2}=400mL\\P_{2}=2atm$

So, isolating the new temperature and substituting the given information, we have:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}\\\\T_{2}=P_{2}V{2}*\frac{T_{1}}{P_{1}V_{1}} \\\\T_{2}=2atm*400mL*\frac{37\°C}{4atm*200mL}=37\°C$

Hence, we have that the temperature will not change because both pressure and volume decreased and increased proportionally, creating the same relationship that we had before the experiment started.

The temperature will be the same, 37°C

Have a nice day!

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3 years ago

A 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, what

ludmilkaskok [199]

The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C

<h3>How to calculate temperature?</h3>

The initial temperature of the copper metal can be calculated using the following formula on calorimetry:

Q = mc∆T

mc∆T (water) = - mc∆T (metal)

Where;

m = mass
c = specific heat capacity
∆T = change in temperature

According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:

400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)

30,096 = 93.6T - 2246.4

93.6T = 32342.4

T = 345.5°C

Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.

Learn more about temperature at: brainly.com/question/15267055

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2 years ago