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Feliz [49]
3 years ago
14

What type of bond is SH2, CO2, SO3? thank you

Chemistry
1 answer:
kykrilka [37]3 years ago
6 0

Answer: SO2(Sulfur Dioxide) Molecular Geometry & Lewis Structure

Explanation: So2Molecular Geometry

We know that the shape which minimizes the repulsions of electronics pairs is adopted by the molecule to form the structure. The molecular shape of SO2 is same as the molecular geometry of Carbon Dioxide (CO2). We will show the bonding of SO2 without making assumption below.

O =S =O

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Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actua
GrogVix [38]

The value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

The rust forms when 4.85X10³ kJ of heat is released is 888.916 g.

<h3>Chemical reaction:</h3>

4 Fe + 3O2 ------ 2Fe2O3

∆H = -1.65×10³kJ

A) Given,

mass of iron = 0.250kg = 250 g

<h3>Calculation of number of moles</h3>

moles = given mass/ molar mass

= 250/ 55.85 g/mol.

= 4.476 mol

As we know that,

For the rusting of 4 moles of Fe, ∆H = -1.65×10³kJ

For the rusting of 4.476 moles of Fe ∆H required can be calculated as

-1.65×10³kJ × 4.476 mol/ 4mol

∆H required = -1.846 × 10³kJ

Now,

when 2 mol of Fe2O3 formed, ∆H = - 1.65×10³kJ

It can be said that,

-1.65×10³kJ energy released when 2 mol of Fe2O3 formed

So, -4.6 × 10³kJ energy released when 2 mol of Fe2O3 formed

= 2 × -4.6 × 10³kJ / -1.65×10³kJ

= 5.57 mol of Fe2O3 formed

Now,

mass of Fe2O3 formed = 5.57 mol × 159.59 g/mol

= 888.916 g

Thus, we calculated that the rust forms when 4.85X10³ kJ of heat is released is 888.916 g. and the value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

learn more about ∆H:

brainly.com/question/24170335

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DISCLAIMER:

The given question is incomplete. Below is the complete question

QUESTION:

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actual process requires water, but a simplified equation is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1.65×10³kJ

a) What is the ∆H when 0.250kg iron rusts.

(b) How much rust forms when 4.85X10³ kJ of heat is released?

7 0
2 years ago
A first order reaction has a rate constant of 0. 543 at 25°C. Given that the activation energy is 75. 9 kj/mol. Calculate the ra
Rasek [7]

The rate constant of first order reaction at 32. 3 °C is 0.343 /s must be less the 0. 543 at 25°C.

First-order reactions are very commonplace. we have already encountered  examples of first-order reactions: the hydrolysis of aspirin and the reaction of t-butyl bromide with water to present t-butanol. every other reaction that famous obvious first-order kinetics is the hydrolysis of the anticancer drug cisplatin.

The value of ok suggests the equilibrium ratio of products to reactants. In an equilibrium combination both reactants and merchandise co-exist. big ok > 1 merchandise are k = 1 neither reactants nor products are desired.

Rate constant K₁  = 0. 543 /s

T₁  = 25°C

Activation energy Eₐ =  75. 9 k j/mol.

T₂ = 32. 3 °C.

K₂ =?

formula;

log K₂/K₁=  Eₐ /2.303 R [1/T₁ - 1/T₂]

putting the value in the equation  

K₂ = 0.343 /s

Hence, The rate constant of first order reaction at 32. 3 °C is 0.343 /s

The specific rate steady is the proportionality consistent touching on the fee of the reaction to the concentrations of reactants. The fee law and the specific charge consistent for any chemical reaction should be determined experimentally. The cost of the charge steady is temperature established.

Learn more about activation energy here:- brainly.com/question/26724488

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4 0
2 years ago
The standard free energy of activation of one reaction A is 95.00 kJ mol–1 (22.71 kcal mol–1). The standard free energy of activ
diamong [38]

Answer:

The answer to the questions are as follows

Reaction B is 4426.28 times faster than reaction A

(b) Reaction B is faster.

Explanation:

To solve the question we are meant to compare both reactions to see which one is faster

The values of the given activation energies are as follows

For A

Ea = 95.00 kJ mol–1 (22.71 kcal mol–1) and

for  B

Ea = 74.20 kJ mol–1 (17.73 kcal mol–1)

T is the same for both reactions and is equal to 298 k

Concentration of both reaction = 1M

The Arrhenius Law is given by

k = Ae^{\frac{-E_{a} }{RT} }

Where

k = rate constant

Ea = activation energy

R = universal gas constant

T = temperature  (Kelvin )

A = Arrhenius factor

Therefore

For reaction A, the rate constant k₁ is given by k₁ = Ae^{\frac{-95000}{(8.314)(298)} }

And for B the rate constant k₂ is given by k₂ = Ae^{\frac{-74200 }{(8.314)(298)} }

k₁ = A×2.225×10⁻¹⁷

k₂ = A×9.850×10⁻¹⁴

As seen from the above Reaction B is faster than reaction A by (A×9.850×10⁻¹⁴)/(A×2.225×10⁻¹⁷) or 4426.28 times

3 0
3 years ago
Please help ASAP
marta [7]

Answer:

0.1

Explanation:

We must first put down the equation of the reaction in order to guide our solution of the question.

2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)

Now from the question, the following were given;

Concentration of acid CA= ??????

Concentration of base CB= 0.299M

Volume of acid VA= 17.8ml

Volume of base VB= 24.7ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CA= CBVBNA/VANB

SUBSTITUTING VALUES;

CA= 0.299 × 24.7 ×2 / 17.8×1

CA= 0.8298 M

But;

pH= -log[H^+]

[H^+] = 0.8298 M

pH= -log[0.8298 M]

pH= 0.1

5 0
4 years ago
What are two ways that substances can pass through a cell membrane out of the cell
Paha777 [63]

Answer:

Passive transport, and Active transport

Explanation:

Do u need an explanation?

If so you can tell me and I'll edit it

Hope this helped a little!

3 0
3 years ago
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