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3 years ago

How many grams of H2 are needed to produce 13.33g of NH3?

1 answer:

3 0

The reaction to form NH3 is : N2 + 3H2-> 2NH3 12,33g NH3 is 12,33/17,03=0,3 =0,724 moles of NH3 moles NH3. So you need 1,5*0,724 = 1,086 moles H2 1,086*2,016 = 2,189 g of H2 is needed ro form 12,33 g NH3

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Draw the structure of the neutral product formed in the reaction of dimethyl malonate and methyl vinyl ketone.

Dmitriy789 [7]

Answer:

Explanation:

The reaction between dimethyl malonate which is an active methylene group with an (∝, β-unsaturated carbonyl compound) i.e methyl vinyl ketone is known as a Micheal Addition reaction. The reaction mechanism starts with the base attack on the β-carbon to remove the acidic ∝-hydrogens and form a carbanion. The carbanion formed(enolate ion) attacks the methyl vinyl ketone(i.e. a nucleophilic attack at the β-carbon) to give a Micheal addition product, this is followed by the protonation to give the neutral product.

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2 years ago

Two easy uses of mixture

lyudmila [28]

Explanation:

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2 years ago

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Sulfuric acid (H₂SO₄), is formed when sulfur dioxide (SO₂) reacts with oxygen and water. Using the balanced chemical reaction ab

Lynna [10]

Answer:

8.5 mol H₂SO₄

Explanation:

It seems the balanced reaction the problem is referring to is absent, however the description matches the following balanced reaction:

2SO₂ + O₂ + 2H₂O → 2H₂SO₄

Now we <u>can convert 8.5 moles of SO₂ into moles of H₂SO₄</u>, using <em>the stoichiometric coefficients of the balanced reaction</em>:

8.5 mol SO₂ * $\frac{2molH_2SO_4}{2molSO_2}$ = 8.5 mol H₂SO₄

5 0

2 years ago

3. What mass of copper could be deposited from a copper(II) Sulphate solution using a current of 5.0 A over 100 seconds? ( F =96

arsen [322]

Mass of copper : 0.165 g

<h3>Further explanation</h3>

Given

5.0 A over 100 seconds

Required

Mass of copper

Solution

Faraday's law:

<em>The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis</em>

<em /> $\tt W=\dfrac{e.i.t}{96500}$ <em />

e = Ar / valence = eqivalent weight

i = current

t = time

W = weight

CuSO₄ ----> Cu²⁺ + SO₄²⁻

Cu ----> Cu²⁺ + 2e

e = Ar/2

= 63,5/2 = 31,75

$\tt W=\dfrac{31.75\times 5\times 100}{96500}=0.165~g$

8 0

2 years ago

A material has an ASTM grain size number of 7. Determine the magnification, if the number of grains per square inch observed is:

Luden [163]

Answer:

A) M = 100X

B) M = 36X

C) M = 178.88X

Explanation:

Given data:

ASTM grain size number 7

a) total grain per inch^2 - 64 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have

$64 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 100 X

B) total grain per inch^2 = 500 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have $500 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 36 X

C) Total grain per inch^2 = 20 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have $20 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 178.88 X

8 0

2 years ago