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jonny [76]
3 years ago
10

When modeling a scientific process, it is more important to organize the parts in a way that makes sense to you than to list the

m sequentially. Please select the best answer from the choices provided T F
Chemistry
2 answers:
AnnyKZ [126]3 years ago
7 0

The answer would actually be false. I just took the test.

WITCHER [35]3 years ago
4 0

The answer is False.

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Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241
Mazyrski [523]

Answer:

-767,2kJ

Explanation:

It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:

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

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ

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

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ

The sum of (4) - (2) produce:

6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ

(6) + 4×(3):

7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ

(7) - 2×(1):

8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ

(8) - 2×(5):

9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= <em>-767,2kJ</em>

I hope it helps!

6 0
3 years ago
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
Ket [755]

Answer:- 10 L of ethane.

Solution:- The given balanced equation is:

2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

From this equation, ethane and oxygen react in 2:7 mol ratio, the ratio of volumes would also be same if they are at same temperature and pressure.

Since 14 L of each gas are taken, the oxygen will be the limiting reactant and ethane will be the excess reactant. Let's calculate the volume of ethane used:

14LO_2(\frac{2LC_2H_6}{7LO_2})

= 4LC_2H_6

From above calculations, 4 L of ethane are used. So, excess volume of ethane left after the completion of reaction = 14 L - 4 L = 10 L

Hence, 10 L of ethane will be remaining.

5 0
3 years ago
How many grams of a stock solution that is 87.5 percent H2SO4 by mass would be needed to make 275 grams of a 55.0 percent by mas
dalvyx [7]
87+55=142
5+0=5
The answer is 142.5
8 0
3 years ago
What is the molarity of a sodium hydroxide solution containing 2 moles of NaOH with a volume of .5L?
Maksim231197 [3]

Answer:

0.4M NaOH

Explanation:

Molarity, M, is an unit of concentration widely used defined as the ratio between moles of solute (In this case, NaOH) and volume of solution in liters.

As the solution contains 2 moles of NaOH-Moles of solute- in 5L of solution, the molarity is:

2 moles NaOH /  5L =

<h3>0.4M NaOH</h3>
5 0
3 years ago
If mercury (Hg) and oxygen (O2) were reacted to form mercury oxide how many molecules of each reactant and product would be need
dimulka [17.4K]
In order to get HgO you would need 2Hg+1O2=2HgO. Since oxygen is diatomic you need two when it stands alone causing you to need two mercuries to balance out the reactants and the product I hope this helps
5 0
3 years ago
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