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4 years ago

Q3. A bridge is built without expansion gaps.

Chemistry

2 answers:

goldenfox [79]4 years ago

7 0

Are you learning about atoms?

marshall27 [118]4 years ago

5 0

Answer:

Q3. A bridge is built without expansion gaps.

Explain what could happen to the bridge if the temperature became:

a) much hotter than the day it was built.

b) much colder than the day it was built.

P.s. (this is not a select option question, explain each according to the question)

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For the following reaction: Fe2O3 + 3CO --> 2Fe + 3CO2 how many grams of Fe2O3 is needed to produce 111 g of Fe?

Afina-wow [57]

You need 158.70 grams of Fe2O3 to produce 111 grams of Fe. This is calculated by using the molar masses and stoichiometric relationship of the two compounds.

Solution:

MM Fe = 55.845 g/mol
MM Fe2O3 = 159.69 g/mol
Fe: Fe2O3 = 2 mol:1 mol

11 g FE (1 mol Fe/55.845 g Fe) (1 mol Fe2O3/2 mol Fe) (159.69 g Fe2O3 / 1 mole Fe2O3) = 158.70 grams Fe2O3

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3 years ago

The unburned, unfiltered fuel parts that remain to be removed through the exhaust are called

Anna007 [38]

The answer to this is hydrocarbons

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3 years ago

Identify the type of solid for FeF3

salantis [7]

Answer:

it is iron and flourine gas

Explanation:

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3 years ago

The incomplete table below shows selected properties of compounds that have ionic, covalent, or metallic bonds.

pogonyaev

Answer:

is there an option for silver? if so, silver is the answer.

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2 years ago

Decide which of the following statements are True and which are False about equilibrium systems:A large value of K means the equ

ivanzaharov [21]

Answer:

a. True

b. False

c. True

d. False

e. False

Explanation:

A. (true) The equilibrium constant K is defined as

$\frac{Products}{reagents}$

In any case

aA +Bb ⇌ Cd +dD

where K is:

$K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

A large value on K means that the concentration of products is bigger than the concentrations of reagents, so the forward reaction is favored, and the equilibrium lies to the right.

B. (False) When we work with gases, we use partial pressure to make calculations in the equilibrium, so we estimate Kp as:

$Kp= \frac{(P_{C})^{c}(P_{D})^{d}}{(P_{A})^{a}(P_{B})^{b}}$

Using the ideal gas law, we can get a relationship between K and Kp

Pv=nRT where $P=\frac{n}{v}*RT$ we know that $\frac{n}{v}$ is the molar concentration. When we replace P in the expression for Kp we get:

$Kp= \frac{[C]^{c}*(RT)^{c}[D]^{d}*(RT)^{d}}{[A]^{a}*(RT)^{a}[B]^{b}*(RT)^{b}}$

Reorganizing the equation:

$Kp= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}*\frac{(RT)^{c+d}}{(RT)^{a+b}}$

We can see K in the expression

$Kp= K*(RT)^{c+d-a-b}$

Delta n = c+d-a-b

$Kp= K*(RT)^{delta n}$

For the reaction

$H_{2}(g) + F_{2}(g)-- equilibrium---2HF(g)$

Delta n = 2-1-1=0

$Kp= K*(RT)^{0}$

So Kp=K in this case.

C. (true) The value of K just depends on the temperature that’s why changing the among of products won’t have any effect on its value.

D. (false) as we can see this reaction involve a heterogeneous system with solids and gases. For convention the concentration for solids and liquids can be considered constant during the reaction that’s why they’re not include in the calculation for the equilibrium constant. Taking this into account the expression for the equilibrium for this reaction is:

$CaCO_{3}(s)---equilibrium----CaO(s) + CO_{2}(g)$

$K= [CO_{2}]$

So we can see that $[CaCO_{3}]$ is not include in the expression.

E. (False) The equilibrium is defined as the point where the rate of the forward reaction is the same to the rate of the reverse reaction. The value of K is telling you which reaction is favored but the rate of both reactions is the same in this point. (see picture)

3 0

3 years ago