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3 years ago

Molecular iodine,

Chemistry

1 answer:

My name is Ann [436]3 years ago

8 0

Answer:

Half-life for this reaction is 2.56s

Explanation:

The general expression in a reaction that follows first-order is:

Ln[A] = -kt + ln[A]₀

Where [A] is concentration of reactant after time t,

k is rate constant = 0.271s⁻¹

[A]₀ is initial concentration of reactant.

Half-life is defined as the time required to decrease the initial concentration of the reactant (I2 in this case) halved.

If [A]₀ = 1

[A] = 1/2

Solving the equation:

Ln[1/2] = -0.271s⁻¹*t + ln[1]

Ln[1/2] = -0.271s⁻¹*t + 0

Ln[1/2] = -0.271s⁻¹t

Ln 2 = 0.271s⁻¹

2.56s = t

Half-life for this reaction is 2.56s

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Which air pollutant contributes to asthma?

Mnenie [13.5K]

The correct answer is the second option. Particulate matter is the air pollutant known to affect people with asthma. It also can cause other illnesses like cough, chest discomfort and a burning sensation in the lungs.

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3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

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4 years ago

1.What are pollen grain? 2. what is an embryo?

swat32

Answer: 2= baby being developed= embryo 1 i dont know sorry

Explanation:

3 0

4 years ago

Read 2 more answers

Please Help! I will give a Brainliest and 18 points! Please do not give me a random, gibberish answer or I will report you and g

3241004551 [841]

Answer:

V2 = 600ml

Explanation:

dilution principle formula

M1V1 = M2V2

C1V1 = C2V2

3 X 20 = 0.1 x V2

60 = 0.1 x V2

V2 = 60/0.1

V2 = 600ml

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5 0

3 years ago

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Which element most likely has the same number of valence electrons as silicon(si)?

Blizzard [7]

Answer:

Carbon, germanium, tin and lead.

Explanation:

The silicon is belong to the carbon family. There are five elements in carbon family carbon, silicon, germanium, tin and lead. These five elements are present in same group i.e group fourteen. The elements present in same group have same number of valance electrons.

For example.

Carbon electronic configuration:

C₆ = [He] 2s² 2p²

Silicon electronic configuration:

Si₁₄ = [Ne] 3s² 3p²

Germanium electronic configuration:

Ge₃₂ = [Ar] 3d¹⁰ 4s² 4p²

Tin electronic configuration:

Sn₅₀ = [Kr] 4d¹⁰ 5s² 5p²

Lead electronic configuration:

Pb₈₂ = [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p²

we can see that in case of all elements there are four valance electrons, which are equal to the valance electrons of silicon.

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3 years ago