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MissTica
3 years ago
6

Which order is correct in listing the bond lengths from shortest to longest? A. single, double, triple B. triple, double, single

C. single, triple, double, D. double, triple, single
Chemistry
1 answer:
Alex787 [66]3 years ago
3 0
A I believe would be the answer
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The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t
ICE Princess25 [194]

Answer:

42,3g of H₂O

Explanation:

For the reaction:

4NH₃(g) + 5O₂(g) ⟶ 4NO(g) + 6H₂O(g)

62,8 g of NH₃ are:

62,8g×(1mol/17,031g) = <em>3,69 moles of NH₃</em>

62,8 g of O₂ are:

62,8g×(1mol/32g) =<em> 1,96 moles of O₂</em>

For a complete reaction of 1,96 moles of O₂ you need:

1,96mol O₂×(4mol NH₃ / 5molO₂) = 1,57 moles NH₃. As you have 3,69 moles, limiting reactant is O₂.

Assuming a complete reaction, 1,96mol O₂ produce:

1,96mol O₂×(6mol H₂O / 5molO₂) = 2,35 moles of H₂O. In grams:

2,35 moles of H₂O×(18,01g/1mol) = <em>42,3g of H₂O</em>

I hope it helps!

3 0
3 years ago
List the types of electricity
Lilit [14]
There are 2 types of electricity, static electricity and current electricity
7 0
2 years ago
Read 2 more answers
15. What volume of water must be added to 300 mL of 0.75 M HCl to dilute the solution to<br> 0.25 M?
Sliva [168]

Known :

V1 = 300 mL

M1 = 0.75 M

M2 = 0.25 M

Solution :

M1 • V1 = M2 • V2

(0.75 M) • (300 mL) = (0.25 M) V2

V2 = 900 mL

Water add to this solution is :

∆V = V2 - V1

∆V = 900 - 300

∆V = 600 mL

7 0
2 years ago
Lewis structures cannot
mariarad [96]
Lewis structure cannot show the numbers of valences
6 0
3 years ago
A 72.0-gram piece of metal at 96.0 °C is placed in 130.0 g of water in a calorimeter at 25.5 °C. The final temperature in the ca
tekilochka [14]

Answer: The answer is S = 0.1528 cal/g °C

Explanation:

By the law of conservation of energy, energy is neither created nor destroyed.

So, energy lost by metal pieces is equal to the energy gained by water in the calorimeter.

Specific heat of water is 1 cal/g °C

⇒ heat energy  Q = mSΔT, where m = mass of a substance

                                                        S = specific heat

                                                        ΔT = change in temperature

Now, the heat lost by metal piece, Q = 72×S×(96-31)

                                                      = 4680×S cal

Heat gained by water, Q = 130×1×(31-25.5)

                                        = 715 cal

⇒ 4680×S = 715.

⇒ S = 0.1528 cal/g °C.

6 0
3 years ago
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