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3 years ago

What type of reaction occurs? S(s) + O2 (g) → SO2(g)

Chemistry

2 answers:

klasskru [66]3 years ago

6 0

Answer:

These reagents are also called as reductants. Redox reactions: Reactions which involve change in oxidation number of the interacting species. Problem 8.4 Justify that the reaction: 2Cu2O(s) + Cu2S(s) → 6Cu(s) + SO2(g) is a redox reaction.

Inessa05 [86]3 years ago

5 0

Answer:

reductants or redox reactions

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For each system listed in the first column of the table below, decide (if possible) whether the change described in the second c

kiruha [24]

Answer:

For each system listed in the first column of the table below, decide (If possible) whether the change described in the second column will Increase the entropy S of the system, decrease S, or leave S unchanged. If you don't have enough information to decide, check the not enough information" button in the last column.

System As A few grams of water vapor (H,0). The water condenses to a liquid at a not enough The carbon dioxide is heated from -5.0 °C to 13.0°c and is also compressed from a volume of 9.0 1 to a volume of 5.0L A few moles of carbon dioxide (CO2) gas. not enough information -15.0 °C to 86.0 °C while the volume is held constant at 2.0 L A few moles of carbon dioxide (CO,) gas. S > 0 not enough 1 Don't Know

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3 years ago

During an experiment,what is the factor you are testing called

Step2247 [10]

Answer:

the answer is variable

4 0

3 years ago

The work function for metallic potassium is 2.3 eV. Calculate the velocity in km/s for electrons ejected from a metallic potassi

mote1985 [20]

Answer:

932.44 km/s

Explanation:

Given that:

The work function of the magnesium = 2.3 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = $2.3\times 1.6022\times 10^{-19}\ J=3.68506\times 10^{-19}\ J$

Using the equation for photoelectric effect as:

$E=\psi _0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

$\lambda$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

v is the velocity of electron

Given, $\lambda=260\ nm=260\times 10^{-9}\ m$

Thus, applying values as:

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{260\times 10^{-9}}=3.68506\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2$

$3.68506\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{260\times \:10^{-9}}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=7.64538\times 10^{-19}-3.68506\times 10^{-19}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=3.96032462\times 10^{-19}$

$v^2=0.869446\times 10^{12}$

v = 9.3244 × 10⁵ m/s

Also, 1 m = 0.001 km

5 0

3 years ago

The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:

bagirrra123 [75]

Answer:

$12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow 14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)$

Explanation:

The steps of the Ostwald process:

$4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)$

$2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g)$

$3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g)$

Combinning the equations:

$4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)$

$(2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g))*2$

$(3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g))*4/3$

$4 NH_3 (g)+ 4 NO (g)+ 7 O_2 (g) + 4 NO_2 (g) +4/3 H_2O (l) \longrightarrow 4 NO (g) + 6 H_2O (g) + 4 NO_2(g) + 8/3 HNO_3 (g) + 4/3 NO (g)$

Simplifying:

$4 NH_3 (g)+ 7 O_2 (g) + \longrightarrow 14/3 H_2O (l) + 8/3 HNO_3 (g) + 4/3 NO (g)$

$12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow 14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)$

The overall reaction is endothermic becuase the formation of new chemical bonds requires energy consumption.

4 0

3 years ago

How many molecules of oxygen are produced when a sample of 38.9 g of water is decomposed by electricity?

statuscvo [17]

Answer:

A) $6.5\times 10^{23}\ \text{molecules}$

Explanation:

m = Mass of water = 38.9

M = Molar mass of water = 18 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

The reaction of electrolysis would be

$2H_2O(l)\rightarrow 2H_2(g)+O_2(g)$

Number of moles of $H_2O$

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{38.9}{18}\ \text{mol}$

From the reaction it can be seen that 2 moles of $H_2O$ gives 1 mole of $O_2$

So, number of moles of $O_2$ produced is

$\dfrac{38.9}{18}\times \dfrac{1}{2}=1.081\ \text{mol}$

Number of molecules

$1.081N_A=1.081\times 6.022\times 10^{23}\\ =6.5\times 10^{23}$

So, $6.5\times 10^{23}\ \text{molecules}$ of oxygen is produced.

8 0

3 years ago