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4 years ago

Which type of reaction is Mg + S → MgS?

Chemistry

2 answers:

Aleksandr [31]4 years ago

8 0

The type of reaction that is Mg +S→ MgS is a synthesis reaction

Explanation

Synthesis reaction as refers to as direct combination is a reaction in which two or more chemical species combine to form more complex product.

The reaction of Mg +S → MgS is a synthesis reaction because;

Mg combine with S to form a more complex product MgS

kkurt [141]4 years ago

4 0

Answer: The correct answer is Option B.

Explanation:

Polymerization reaction is defined as a reaction in which monomer molecules react together to form a polymer chains which are three-dimensional networks.

Synthesis reaction is defined as a chemical reaction in which two or more chemical species react in their naturally occurring state to form a single large compound.

$A+B\rightarrow AB$

Replacement reaction is defined as a chemical reaction in which exchange of ions takes place. They are of two types: Single replacement and double replacement.

$A+BC\rightarrow AC+B\\\\AB+CD\rightarrow AD+CB$

Decomposition reaction is a type of chemical reaction in which a larger molecule breaks down into two or more smaller molecules.

$AB\rightarrow A+B$

The given reaction follows:

$Mg(s)+S(s)\rightarrow MgS(s)$

This is a type of synthesis reaction.

Hence, the correct answer is Option B.

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3 years ago

A 25 mL sample of 0.100 M HNO3 completely reacts with NaOH according to this equation:

Anastasy [175]

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3 years ago

Read 2 more answers

Please help this is due tomorrow!

lbvjy [14]

I had this same question on one of are worksheets is D I can’t really explain it but the answer is D

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4 years ago

The rate of a reaction is measured by how fast a reactant is used up to how fast a product is formed.

bezimeni [28]

A. True.

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4 years ago

Read 2 more answers

A. The reactant concentration in a zero-order reaction was 8.00×10−2 M after 155s and 3.00×10−2 M after 355s . What is the rate

irga5000 [103]

Answer:

A) The rate constant is 2.50 × 10⁻⁴ M/s.

B) The initial concentration of the reactant is 11.9 × 10⁻² M.

C) The rate constant is 0.0525 s⁻¹

D) The rate constant is 0.0294 M⁻¹ s⁻¹

Explanation:

Hi there!

A) The equation for a zero-order reaction is the following:

[A] = -kt + [A₀]

Where:

[A] = concentrationo f reactant A at time t.

[A₀] = initial concentration of reactant A.

t = time.

k = rate constant.

We know that at t = 155 s, [A] = 8.00 × 10⁻² M and at t = 355 s [A] = 3.00 × 10⁻² M. Then:

8.00 × 10⁻² M = -k (155 s) + [A₀]

3.00 × 10⁻² M = -k (355 s) + [A₀]

We have a system of 2 equations with 2 unknowns, let´s solve it!

Let´s solve the first equation for [A₀]:

8.00 × 10⁻² M = -k (155 s) + [A₀]

8.00 × 10⁻² M + 155 s · k = [A₀]

Replacing [A₀] in the second equation:

3.00 × 10⁻² M = -k (355 s) + [A₀]

3.00 × 10⁻² M = -k (355 s) + 8.00 × 10⁻² M + 155 s · k

3.00 × 10⁻² M - 8.00 × 10⁻² M = -355 s · k + 155 s · k

-5.00 × 10⁻² M = -200 s · k

-5.00 × 10⁻² M/ -200 s = k

k = 2.50 × 10⁻⁴ M/s

The rate constant is 2.50 × 10⁻⁴ M/s

B) The initial reactant conentration will be:

8.00 × 10⁻² M + 155 s · k = [A₀]

8.00 × 10⁻² M + 155 s · 2.50 × 10⁻⁴ M/s = [A₀]

[A₀] = 11.9 × 10⁻² M

The initial concentration of the reactant is 11.9 × 10⁻² M

C) In this case, the equation is the following:

ln[A] = -kt + ln([A₀])

Then:

ln(7.60 × 10⁻² M) = -35.0 s · k + ln([A₀])

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln([A₀])

Let´s solve the first equation for ln([A₀]) and replace it in the second equation:

ln(7.60 × 10⁻² M) = -35.0 s · k + ln([A₀])

ln(7.60 × 10⁻² M) + 35.0 s · k = ln([A₀]

Replacing ln([A₀]) in the second equation:

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln([A₀])

ln(5.50 × 10⁻³ M) = -85.0 s · k + ln(7.60 × 10⁻² M) + 35.0 s · k

ln(5.50 × 10⁻³ M) - ln(7.60 × 10⁻² M) = -85.0 s · k + 35.0 s · k

ln(5.50 × 10⁻³ M) - ln(7.60 × 10⁻² M) = -50.0 s · k

(ln(5.50 × 10⁻³ M) - ln(7.60 × 10⁻² M)) / -50.0 s = k

k = 0.0525 s⁻¹

The rate constant is 0.0525 s⁻¹

D) In a second order reaction, the equation is as follows:

1/[A] = 1/[A₀] + kt

Then, we have the following system of equations:

1/ 0.510 M = 1/[A₀] + 205 s · k

1/5.10 × 10⁻² M = 1/[A₀] + 805 s · k

Let´s solve the first equation for 1/[A₀]:

1/ 0.510 M = 1/[A₀] + 205 s · k

1/ 0.510 M - 205 s · k = 1/[A₀]

Now let´s replace 1/[A₀] in the second equation:

1/5.10 × 10⁻² M = 1/[A₀] + 805 s · k

1/5.10 × 10⁻² M = 1/ 0.510 M - 205 s · k + 805 s · k

1/5.10 × 10⁻² M - 1/ 0.510 M = - 205 s · k + 805 s · k

1/5.10 × 10⁻² M - 1/ 0.510 M = 600 s · k

(1/5.10 × 10⁻² M - 1/ 0.510 M)/ 600 s = k

k = 0.0294 M⁻¹ s⁻¹

The rate constant is 0.0294 M⁻¹ s⁻¹

8 0

3 years ago