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3 years ago

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If the reaction below takes place at STP how many L of ammonia (NH3) will be formed if one starts with 3.45 moles of hydrogen an

d an excess of nitrogen? Be sure to balance the equation before you begin and round to the proper accuracy. Please leave any units off.

1 answer:

lesya692 [45]3 years ago

6 0

Answer:

51.5 L of ammonia

Explanation:

Ammonia is produced according to this equation:

N₂(g) + 3H₂(g) → 2NH₃ (g)

Let's propose this rule of three:

3 moles of hydrogen can produce 2 moles of ammonia

Then 3.45 moles of hydrogen, will produce (3.45 . 2) / 3 = 2.3 moles of NH₃

We apply the Ideal Gases Law at STP to determine the volume

STP = 1 atm and 273K

1 atm . V = 2.3 mol . 0.082 . 273K

V = (2.3 mol . 0.082 . 273K) / 1 atm = 51.5 L

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Calculate the moles of SO₂ produced when 5 moles of FeS₂ reacts according to the equation: 4FeS₂+ 11O₂ → 2Fe₂O₃ + 8SO₂

Gala2k [10]

Answer:

10 moles of SO₂ are produced when 5 moles of FeS₂

Explanation:

Stoichiometry: it is the theoretical proportion in which the chemical species are combined in a chemical reaction. The stoichiometric equation of a chemical reaction relates molecules or number of moles of all the reagents and products that participate in the reaction.

In other words, stoichiometry establishes relationships between the molecules or elements that make up the reactants of a chemical equation with the products of said reaction. The relationships established are molar relationships (that is, moles) between the compounds or elements that make up the chemical equation.

The stoichiometric coefficients of a chemical reaction indicate the proportion in which said substances react.

Taking into account the above, you can apply the following rule of three: by stoichiometry if 4 moles of FeS₂ produce 8 moles of SO₂, then when reacting 5 moles of FeS₂ how many moles of SO₂ will they produce?

$moles of SO_{2} =\frac{5 moles of FeS_{2} *8moles of SO_{2}}{4 moles of FeS_{2}}$

moles of SO₂= 10

<u><em>10 moles of SO₂ are produced when 5 moles of FeS₂</em></u>

7 0

3 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago

According to the cup wall diagrams, why does the double wall vacuum

sveta [45]

Answer:

i do not now

Explanation:

8 0

3 years ago

Solution of the Schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the behav

kvv77 [185]

Answer :

'n' specifies → (B) The energy and average distance from the nucleus.

'l' specifies → (C) The subshell orbital shape.

'ml' specifies → (A) The orbital orientation.

Explanation :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as $m_l$ . The value of this quantum number ranges from $(-l\text{ to }+l)$ . When l = 2, the value of

Spin Quantum number : It describes the direction of electron spin. This is represented as $m_s$ . The value of this is $+\frac{1}{2}$ for upward spin and $-\frac{1}{2}$ for downward spin.

As per question we conclude that,

'n' specifies → The energy and average distance from the nucleus.

'l' specifies → The subshell orbital shape.

'ml' specifies → The orbital orientation.

5 0

3 years ago

How many liters of a 4.0 M CaCl2 solution would contain 2 moles of CaCl2?

nekit [7.7K]

Answer:

The answer to your question is 0.5 liters

Explanation:

Data

[CaCl₂] = 4.0 M

number of moles = 2

volume = ?

Process

To solve this problem use the formula of Molarity and solve it for volume, substitute the values and simplify.

-Formula

Molarity = moles / volume

-Solve for volume

Volume = moles / molarity

-Substitution

Volume = 2/4

-Simplification

Volume = 0.5 liters.

5 0

3 years ago