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svp [43]
3 years ago
8

Chemical reactions in which large molecules are built from small molecules are classified as ______ reactions. In contrast, ____

__ reactions involve breaking large molecules down into smaller ones.
Chemistry
1 answer:
Luda [366]3 years ago
6 0
<h2>Exothermic and Endothermic Reaction</h2>

Explanation:

  • Chemical reactions in which large molecules are built from small molecules are classified as exothermic reactions. During these reactions, a huge amount of energy is released.
  • When bonds are formed to attain its stability there is the loss of energy as heat energy.  
  • In contrast, Endothermic reactions involve breaking large molecules down into smaller ones. During this reaction, a large amount of energy is involved.
  • When bonds break to attain its stability there is gain of energy.
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Tuliskan persamaan tetapan kesetimbangan untuk reaksi-reaksi berikut a. Fe3+(aq) + SCN– (aq) ↔ FeSCN3+(aq) b. 3Fe(s) + 4H2O(g) ↔
nevsk [136]

Answer:

a. K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}

b. K_p = \dfrac{[H_2]^4}{[H_2O]^4}

Explanation:

Untuk semua jenis reaksi umum:

aA + bB\iff cC + dD

Konstanta kesetimbangan K_c = \dfrac{[C]^c [D]^d}{[A]^a[B]^b}

Dari pertanyaan yang diberikan:

a. Fe3^+_{(aq)} + SCN^-_{ (aq)} \iff FeSCN^{3+}_{(aq) }

Konstanta kesetimbangan:

K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}

b. 3Fe_{(s)} + 4H_2O_{(g)} \iff Fe_3O_4_{(s)} + 4H_{2(g)}

Konstanta kesetimbangan untuk tekanan parsial K_p

K_p = \dfrac{[H_2]^4}{[H_2O]^4}

Karena Fe3O4 (s) hadir sebagai padatan.

8 0
2 years ago
Which three temperature readings all mean the same thing?
allochka39001 [22]

Answer: The correct answer is D. 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit.

Explanation:

Conversion of degree Celsius to Kelvin :

K=^oC+273

Conversion of degree Celsius to degrees Fahrenheit :

^oF=(\frac{9}{5}\times ^oC)+32

By using these two conversion factors, we get the three temperature readings all mean the same thing.

For option A :

K=^oC+273=100+273=373K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF

For option B :

K=^oC+273=100+273=373K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF

For option C :

K=^oC+273=0+273=273K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF

For option D :

K=^oC+273=0+273=273K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF

From the given options, only option (D) is correct.

Hence, the correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit

Hope this helps!

8 0
3 years ago
Read 2 more answers
During a synthesis reaction, 1.8 grams of magnesium reacted with 6.0 grams of oxygen. What is the maximum amount of magnesium ox
skelet666 [1.2K]

Answer:

2.9 grams.

Explanation:

  • From the balanced reaction:

<em>Mg + 1/2O₂ → MgO,</em>

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

  • We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:

no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.

no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.

<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.074 mol of Mg produce → 0.074 mol of MgO.

<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>

5 0
3 years ago
At 450°C, ammonia gas will decompose according to the following equation: 2 NH3 (g)  N2 (g) + 3 H2 (g) Kc = 4.50 at 475˚C An u
velikii [3]

Answer:

0.2024 M

Explanation:

For the decomposition reactio given, let's do an equilibrium chart. Let's call the initial concentration of NH₃ as C:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

C 0 0 Initial

-2x +x +3x Reacts (stoichiometry is 1:1:3)

C - 2x x 3x Equilibrium

3x = 0.252

x = 0.084 M

The equilibrium constant (Kc) is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of reactants concentrations elevated by their coefficients.

Kc = ([H₂]³*[N₂])/([NH₃]²)

4.50 = [(0.252)³*(0.084)]/(C - 2*0.084)²

4.50 = 0.00533/(C - 0.168)²

4.50 = 0.00533/(C² - 0.336C + 0.028224)

4.50C² - 1.512C + 0.127008 = 0.00533

4.50C² - 1.512C + 0.121678 = 0

Solving the equation by a graphic calculator, for C > 0.168

C = 0.2024 M

4 0
3 years ago
The smallest unit which maintains the physical properties of a compound is a(n _____. atom element molecule alloythe smallest un
LuckyWell [14K]
The answer that would best complete the given statement above is the term ATOM. <span>The smallest unit which maintains the physical properties of a compound is an ATOM. But for a compound, it would be molecule. Hope this answers your question.</span>
5 0
3 years ago
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