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In preparation for a blizzard, a city dispatches a crew to spread salt on the street surface of the bridges around town. The sal

Ilia_Sergeevich [38]

0 degrees to -6 or -16 degrees Celsius

3 0

3 years ago

P4O10(s) = -3110 kJ/mol H2O(l) = -286 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in enthalpy for the following process:

mash [69]

Answer:

-290KJ/mol

Explanation:

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}

ΔHrxn = 4(-1279) - [6(-286) - 3110]

= -5116 -(-1716-3110)

= -5116-(-4826)

= -5116 + 4826 = -290KJ/mol

6 0

2 years ago

Consider the following system at equilibrium:

alexgriva [62]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

6 0

2 years ago

Consider the reaction directly below and answer parts a and b. C6H4(OH)2 (l) + H2O2 (l) à C6H4O2 (l) + 2 H2O (l)

dedylja [7]

Answer:

a. -206,4kJ

b. Surroundings will gain heat.

c. -115kJ are given off.

Explanation:

It is possible to obtain ΔH of a reaction using Hess's law that consist in sum the different ΔH's of other reactions until obtain the reaction you need.

Using:

(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

(2) H₂(g) + O₂(g) → H₂O₂ (l) ΔH: -187.8 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH: -285.8 kJ

It is possible to obtain:

C₆H₄(OH)₂(l) + H₂O₂(l) → C₆H₄O₂(l) + 2H₂O(l)

From (1)-(2)+2×(3). That is:

(1) C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

-(2) H₂O₂(l) → H₂(g) + O₂(g) ΔH: +187.8 kJ

2x(3) 2H₂(g) + O₂(g) → 2H₂O(l) ΔH: 2×-285.8 kJ

The ΔH you obtain is:

+177,4kJ + 187,8kJ - 2×285.8 kJ = -206,4kJ

b. When ΔH of a reaction is <0, the reaction is exothermic, that means that the reaction produce heat and the surroundings will gain this heat.

c. 20,0g of H₂O are:

20,0g× $\frac{1mol}{18,01g}$ = 1,11 mol H₂O

As 2 moles of H₂O are produced when -206,4kJ are given off, when 1,11mol of H₂O are produced, there are given off:

1,11mol H₂O× $\frac{-206,4kJ}{2mol}$ = -115kJ

I hope it helps!

8 0

3 years ago

Which phenomena naturally warms earth lower atmosphere and surface?

Alenkinab [10]

The phenomenon that naturally warms the earth's lower atmosphere and surface is called the greenhouse effect.

Put me as Brainliest!!!

6 0

3 years ago

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