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Vlad [161]
3 years ago
13

PLS HELP!!!!

Chemistry
2 answers:
Mashcka [7]3 years ago
7 0

Answer:

the 3rd answer

White raven [17]3 years ago
4 0

How do the products of chemical reactions compare to their reactants?

Answer:

The products often have completely different properties than the reactants.

You might be interested in
Given the following reactions 2S (s) + 3O2 (g) → 2SO3 (g) ΔH = -790 kJ S (s) + O2 (g) → SO2 (g) ΔH = -297 kJ the enthalpy of the
Trava [24]

Answer:

-196 kJ

Explanation:

By the Hess' Law, the enthalpy of a global reaction is the sum of the enthalpies of the steps reactions. If the reaction is multiplied by a constant, the value of the enthalpy must be multiplied by the same constant, and if the reaction is inverted, the signal of the enthalpy must be inverted too.

2S(s) + 3O₂(g) → 2SO₃(g)  ΔH = -790 kJ

S(s) + O₂(g) → SO₂(g)         ΔH = -297 kJ (inverted and multiplied by 2)

2S(s) + 3O₂(g) → 2SO₃(g)  ΔH = -790 kJ

2SO₂(g) → 2S(s) + 2O₂(g)   ΔH = +594 kJ

-------------------------------------------------------------

2S(s) + 3O₂(g) + 2SO₂(g) → 2SO₃(g) + 2S(s) + 2O₂(g)

Simplifing the compounds that are in both sides (bolded):

2SO₂(g) + O₂(g) → 2SO₃(g) ΔH = -790 + 594 = -196 kJ

4 0
4 years ago
Gas in a balloon occupies 3.3 L. What volume will it occupy if the pressure is changed from 100.0 kPa to 90.0 kPa (at constant t
german
Since the temperature is a constant, we can use Boyle's law to solve this.<span>

<span>Boyle' law says "at a constant temperature, the pressure of a fixed amount of an ideal gas is inversely proportional to its volume.

P α 1/V               </span>⇒                 PV = k (constant)<span>

Where, P is the pressure of the gas and V is the volume.

<span>Here, we assume that the </span>gas in the balloon is an ideal gas. 

We can use Boyle's law for these two situations as,
            P</span>₁V₁ = P₂V₂<span>

P₁ = 100.0 kPa = 1 x 10⁵ Pa
V₁ = 3.3 L
P₂ = 90.0 x 10³ Pa
V₂ =?

By substitution,
    1 x 10⁵ Pa x 3.3 L = 90 x 10³ Pa x V₂</span><span>
                            V</span>₂ = 3.7 L<span>
</span><span>Hence, the volume of gas when pressure is 90.0 kPa is 3.7 L.</span></span>
8 0
3 years ago
A particle that has the electron configuration 1s²2s²2p³ would:
notka56 [123]

Answer:

Have 2 filled orbitals and 3 partially filled orbitals.

Explanation:

Hello there!

In this case, according to the given information of the electron configuration for that particle; it is possible for us to infer it has 5 valence electrons, as the electrons on its outermost shell (2). Moreover, we undertand this particle needs three bonds, does not have neither the electron configuration of a noble gas which ends by p⁶ nor that of an alkali earth metal as it ends by s².

Therefore, we infer the correct answer is Have 2 filled orbitals and 3 partially filled orbitals because according to the Hund's rule, the s orbital is fulfilled and the p orbital has 1 electron orbital fulfilled and two partially filled orbitals.

Regards!

5 0
3 years ago
A substance with a ph of 9 has
Arisa [49]

Answer:

c) more OH⁻ ions than H₃O⁺ ions

Explanation:

A substance with a PH of 9 implies that it has more OH⁻ ions than H₃O⁺ ions.

Such substances are said to be an alkaline or a base.

A base is a substance the produces excess hydroxyl ion in aqueous solutions.

An acid will produce excess hydroxonium ions in a solution.

So, the pH scale is used to indicate whether a substance is an acid or base or non of them.

Acids have pH of less than 7

Bases have pH of > 7

4 0
3 years ago
A 475 cm3 sample of gas at standard temperature and pressure is allowed to expand until it occupies a
Andrej [43]

The final temperature : 345 K

<h3> Further explanation </h3>

Given

475 cm³ initial volume

600 cm³ final volume

Required

The final temperature

Solution

At standard temperature and pressure , T = 273 K and 1 atm

Charles's Law  :

When the gas pressure is kept constant, the gas volume is proportional to the temperature  

V₁/T₁=V₂/T₂

Input the value :

T₂=(V₂T₁)/V₁

T₂=(600 x 273)/475

T₂=345 K

4 0
3 years ago
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