Atmospheric pressure on the peak of Mt. Everest can be as low as

In what way is sedimentary rock formed in nature with a sieve

ella [17]

Sedimentary rock is made of Brocken down rocks and living things.

4 0

3 years ago

PLEASEEE HELPPP!! Due today!!!!

dolphi86 [110]

Answer:

B

Explanation:

A chemical reaction equation is balanced if the number of atoms of each element on the right hand side of the reaction equation is the same as the number of atoms of the same element on the left hand side of the reaction equation.

By inspection, the chemical reaction written in the question is a balanced chemical reaction equation.

However, the statement in option B that more molecules of N2 are needed to produce NH3 than molecules of H2 is a false statement since only one mole of N2 and three moles of H2 are required to produce NH3 according to the reaction equation as written in the question.

3 0

3 years ago

Choose the solvent below that would show the greatest boiling point elevation when used to make a 0.10 m nonelectrolyte solution

postnew [5]

Answer is: 1) ccl4, kb = 29.9°c/m, carbon tetrachloride has the greatest boiling point elevation.

The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.

ΔTb - the boiling point elevation.
Kb - the ebullioscopic constant.
b - molality of the solution.
So the highest boiling poing elevation will be for solution with highest ebullioscopic constant because molality is the same.

5 0

3 years ago

Read 2 more answers

Give the percent yield when 162.8 g of CO2 are formed from the reaction of excess amount of C8H18 and with 218.0 grams of O2. (2

katen-ka-za [31]

Percent yield is 23.11 % when 162.8 g of CO2 are formed from the reaction of excess amount of C8H18 and with 218.0 grams of O2.

Explanation:

Balanced equation for the chemical reaction:

2C8H18 + 25O2 → 16CO2 + 18H2O

data given:

CO2 formed (actual yield) = 162.8 grams

mass of oxygen = 218 grams

16 moles of CO2 formed when 5 moles of oxygen reacted

3.6 moles of CO2 formed when 6.8 moles of oxygen reacted.

In the reaction 16 moles of CO2 will have 44.01 x 16

theoretical yield of CO2 = 704.16 grams

percent yield = $\frac{actual yield}{theoritical yield}$ x100

putting the values in the above equation

percent yield = $\frac{162.8}{704.16}$ x 100

= 0.23 x 100

= 23.11 %

Percent yield is 23.11 %.

7 0

3 years ago

Consider the following system at equilibrium: 2A(aq)+2B(aq)⇌5C(aq) Classify each of the following actions by whether it causes a

Alexandra [31]

Answer:

Rightwardshift: (a), (b), (f) and (h)

Leftwardshift: (c), (d), and (e)

No shift: (g)

Explanation:

1. Balanced chemical equation (given):

$2A(aq)+2B(aq)\rightleftharpoons 5C(aq)$

2. Equilibrium constant

The equilibrium constant is the ratio of the product of the concentrations of the products, at equilibrium, each raised to its stoichiometric coefficient, to the product of the concentrations of the reactants, at the equilibrium, each raised to its stoichiometric coefficient.

$K_{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}$

a. Increase [B]

Rightward shift

Since, by assumption, the temperature of the reaction is the same, the equilibrium constant $K_{c}$ is the same, meaning that an increase in the concentration of the species B must cause a rightward shift to increase the concentration of the species C, such that the ratio expressed by the equilibrium constant remains unchanged.

b. Increase [A]

Rightward shift.

This is exactly the same case for the increase of [B], since it is in the same side of the equilibrium chemical equation.

c. Increase [C]

Leftward shift.

C is on the right side of the equilibrium equation, thus, following Le Chatelier's principle, an increase of its concentration must shift the reaction to the left to restore the equilibrium. Of course, same conclusion is drawn by analyzing the expression for $K_{c}$ : by increasing the denominator the numerator has to increase to keep the same value of $K_{c}$

d. Decrease [A]

Leftward shift.

This is the opposite change to the case {b), thus it will cause the opposite effect.

e. Decrease[B]

Leftward shift.

This is the opposite to case (a), thus it will cause the opposite change.

f. Decrease [C]

Rightward shift.

This is the opposite to case (c), thus it will cause the opposite change.

g. Double [A] and reduce [B] to one half

No shift

You need to perform some calculations and determine the reaction coefficient, $Q_c$ to compare with the equilibrium constant $K_{c}$ .

The expression for $Q_c$ has the same form of the equation for $K_{c}$ . but the it uses the inital concentrations instead of the equilibrium concentrations.

$Q{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}$

Doubling [A] and reducing [B] to one half would leave the product of [A]² by [B]² unchanged, thus $Q_c$ will be equal to $K_{c}$ .

When $Q_c$ = $K_{c}$ the reaction is at equilibrium, so no shift will occur.

h. Double both [B] and [C]

Rightward shift.

Again, using the expression for $Q_c$ , you will realize that the [C] is raised to the fifth power (5) while [B] is squared (power 2). That means that $Q_c$ will be greater than $K_{c}$ ..

When $Q_c$ > $K_{c}$ the equilibrium must be displaced to the left some of the reactants will need to become products, causing the reaction to shift to the right.

Summarizing:

Rightwardshift: (a), (b), (f) and (h)

Leftwardshift: (c), (d), and (e)

No shift: (g)

4 0

4 years ago