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3 years ago

Identify the single displacement reaction.

Chemistry

2 answers:

Maurinko [17]3 years ago

7 0

The displacement reaction is D

pishuonlain [190]3 years ago

3 0

Answer : The correct option is, (D)

Explanation :

Option A reaction : $C_4H_{12}+7O_2\rightarrow 6H_2O+4CO_2$

It is a combustion reaction. A reaction in which a hydrocarbon react with the oxygen to give product as carbon dioxide and water.

Option B reaction : $2H_2+O_2\rightarrow 2H_2O$

It is a combination reaction. A reaction in which the two or more reactants react to give a product.

Option C reaction : $Al_2S_3\rightarrow 2Al+3S$

It is a decomposition reaction. A reaction in which a reactant decomposes to form two or more products.

Option D reaction : $Cl_2+2KBr\rightarrow 2KCl+Br_2$

It is a single displacement reaction. It is a reaction in which the more reactive element displace the less reactive element. In this reaction, most reactive element chlorine displaces the less reactive element bromine.

Hence, the correct option is, (D)

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Write the following number using Scientific notation : 13,215,296.50 *

ludmilkaskok [199]

Answer:

13.22x10^7

Explanation:

7 0

2 years ago

Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an

Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0

3 years ago

Consider the reaction: N2(g) + O2(g) ⇄ 2NO(g) Kc = 0.10 at 2000oC Starting with initial concentrations of 0.040 mol/L of N2 and

IrinaVladis [17]

Answer:

0.011 mol/L

Explanation:

This can be solved with something called an ICE table.

I = initial

C = change

E = equilibrium

Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.

x amount of N₂ reacts. Since the stoichiometry is 1:1, x amount of O₂ also reacts. This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

$\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]$

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

Substituting:

0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )

Solving:

0.10 = (2x)² / (0.04 - x)²

√0.10 = 2x / (0.04 - x)

(√0.10) (0.04 - x) = 2x

(√0.10)(0.04) - (√0.10)x = 2x

(√0.10)(0.04) = 2x + (√0.10)x

(√0.10)(0.04) = (2 + √0.10)x

x = (√0.10)(0.04) / (2 + √0.10)

x = 0.0055

At equilibrium, the concentration of NO is 2x. So the answer is:

[NO] = 2x

[NO] = 0.011

The equilibrium concentration of NO is 0.011 mol/L.

3 0

2 years ago

Why are x-rays telescopes and gamma-ray telescopes in space?

pogonyaev

Because those can leak and cause cancer

5 0

3 years ago

1. Sometimes scientists need to be able to replicate the results of a scientific experiment. Explain in what situation(s) this w

Andrej [43]

I need this also I’m just here for points

4 0

2 years ago