Someone please help me

Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O

Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

ΔH = -793,6 kJ

I hope it helps!

5 0

3 years ago

200.0 mL of 3.85 M HCl is added to 100.0 mL of 4.6 M barium hydroxide. The reaction goes to completion. What is the concentratio

Ede4ka [16]

Answer:

2.387 mol/L

Explanation:

The reaction that takes place is:

2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

First we calculate how many moles of each reagent were added:

HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl

Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂

460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.

Now we calculate how many moles of Ba(OH)₂ reacted, by converting the total number of HCl moles to Ba(OH)₂ moles:

203.85 mmol HCl * $\frac{1mmolBa(OH)_{2}}{2mmolHCl}$ = 101.925 mmol Ba(OH)₂

This means the remaining Ba(OH)₂ is:

460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂

There are two OH⁻ moles per Ba(OH)₂ mol:

OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻

Finally we divide the number of OH⁻ moles by the total volume (100 mL + 200 mL):

716.15 mmol OH⁻ / 300.0 mL = 2.387 M

So the answer is 2.387 mol/L

7 0

3 years ago

The ion with a +3 charge 28 electrons and mass number of 71

Talja [164]

Answer:

Good luck!

Explanation:

6 0

3 years ago

Read 2 more answers

Determine whether each of the following statements about sulfur oxide and/or nitrogen oxide pollution is true or false.

astraxan [27]

Answer:

a) False

b) False

c) True

Explanation:

a) Most sulfur oxides formed during combustion come from sulfur in the air: The sulfur burns in presence of oxygen to produce sulfur oxides, the source of sulfur is anthropogenic activities (Human activities).

So the given statement is false.

b) A flue gas scrubber removes SO2 pollution in a coal-fired power plant by using lime to convert the SO2 into alcohols:

False

The lime reacts with sulfur dioxide to give carbon dioxide and calcium sulfite.

$CaCO_{3}+SO_{2}--->CaSO_{3}+CO_{2}$

c) True

c)

3 0

3 years ago

In the single displacement reaction of HCl and magnesium, how many grams of magnesium chloride would be produced if 1.05g of mag

iren [92.7K]

Answer:

4.16g of MgCl2

Explanation:

First, let us generate a balanced equation for the reaction:

Mg + 2HCl —> MgCl2 + H2

Molar Mass of Mg = 24g/mol

Molar Mass of MgCl2 = 24 + (2x35.5) = 24 + 71 = 95g

From the equation,

24g of Mg produced 95g of MgCl2.

Therefore, 1.05g of Mg will produce = (1.05x95)/24 = 4.16g of MgCl2

5 0

3 years ago