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2 years ago

In which conditions combination and displacement reaction takes place explain with example.

Chemistry

2 answers:

expeople1 [14]2 years ago

8 0

<h2>Answer:</h2>

In <u>Combination reaction</u>, two or more elements combined to form one compound of different properties.

C(s) + O2(g) ⇢ CO2(g).
H2(g) + O2(g) ⇢ H20(l).

In <u>Displacement reation</u>, the high reactive element displaces the low reactive element and formed compound of different properties.

Fe(s) + CuSo4(aq) ⇢ FeSo4(aq) + Cu(s).
AgNO3(aq) + Cu(s) ⇢ CuNO3(aq) + Ag(s).

scZoUnD [109]2 years ago

5 0

Answer:

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In Combination reaction, two or more elements combined to form one compound of different properties.

C(s) + O2(g) ⇢ CO2(g).
H2(g) + O2(g) ⇢ H20(l).

In Displacement reation, the high reactive element displaces the low reactive element and formed compound of different properties.

Fe(s) + CuSo4(aq) ⇢ FeSo4(aq) + Cu(s).
AgNO3(aq) + Cu(s) ⇢ CuNO3(aq) + Ag(s).

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Which element(s) is/are not balanced in this equation?

maks197457 [2]

The answer is d because in both sides there are 1 magnesiums 1 oxygens 2 lithiums and 2 chlorines

6 0

3 years ago

Read 2 more answers

Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.

Delvig [45]

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED'] = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E] (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED] (dissociation constant)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633 * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED'] E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED'] = 0.0579

5 0

2 years ago

How many moles are represented by 3.01 x10^24 oxygen atoms?

asambeis [7]

<h3>Answer:</h3>

5.00 mol O₂

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.<u> </u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.01 × 10²⁴ atoms O₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.01 \cdot 10^{24} \ atoms \ O_2(\frac{1 \ mol \ O_2}{6.022 \cdot 10^{23} \ atoms \ O_2})$
Multiply/Divide: $\displaystyle 4.99834 \ mol \ O_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

4.99834 mol O₂ ≈ 5.00 mol O₂

4 0

2 years ago

PLEASE HELP ME!!! ASAP

shtirl [24]

Answer:

Theoretical yield of the reaction = 34 g

Excess reactant is hydrogen

Limiting reactant is nitrogen

Explanation:

Given there is 100 g of nitrogen and 100 g of hydrogen

Number of moles of nitrogen = 100 ÷ 28 = 3·57

Number of moles of hydrogen = 100 ÷ 2 = 50

Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation

N2 + 3H2 → 2NH3

From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed

Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen

⇒ We require 10·71 moles of hydrogen

But we have 50 moles of hydrogen

∴ Limiting reactant is nitrogen and excess reactant is hydrogen

From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia

Molecular weight of ammonia = 17 g

∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g

5 0

3 years ago

Balanced equation for solid magnesium oxide

Y_Kistochka [10]

Answer:

Mg + O2 → MgO

Explanation:

3 0

2 years ago

In which conditions combination and displacement reaction takes place explain with example.​

In which conditions combination and displacement reaction takes place explain with example.