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2 years ago

What volume of o2 is needed to react fully with 720. Ml of nh3

1 answer:

6 0

Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.

<h3>What is volume?</h3>

Volume is the area occupied by the substance and is the ratio of the mass to the density.

At STP, 1 mole of gas occupies 22.4 L of volume

Given,

Volume of ammonia reacted = 0.720 L

The combustion reaction is shown as,

$\rm 4NH_{3} + 5O_{2} \rightarrow 4NO +6H_{2}O$

From the stoichiometry of the reaction, it can be said that,

$(4 \times 22.4)$ L of ammonia reacts with $(5 \times 22.4)$ L of oxygen gas.

So, 0.720 L of ammonia will react with:

$\dfrac{ (5 \times 22.4)}{(4 \times 22.4)} \times 0.720 = 0.9 \;\rm L$

Therefore, the volume of oxygen required is 900 mL.

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#12 PLEASE HELP IM TAKING A TIMED TEST

Goshia [24]

Answer:

Option B

Explanation:

For the problem in question 12, option B is the correct choice. It rightly expresses how electrons will fill the third energy level of aluminum.

Aluminum is an element with 13 electrons. It has 3 energy levels.

The shell notation is given as 2 8 3

Sublevel notation = 1s² 2s² 2p⁶ 3s² 3p¹

The third energy level is given as 3s² 3p¹

We can represent it as: ⇅ ↑

3s² 3p¹

3 0

3 years ago

List 2 everyday examples for<br> decanting <br> centrifuge <br> filtration

AfilCa [17]

Answer:decanting is done when decanting the liquid before using mustard in order to make a less messy hotdog,when frying hamburger meat for spaghetti source you decant the grease from the pan

filtration is used in heating,ventilation,air conditioning systems at home in order to remove particles and somites odours and chemicals

Explanation:

7 0

4 years ago

If the compound contains 35.06 % cl by mass, what is the identity of the metal?

Setler [38]

Answer is: identity of the metal is gold (Au).

ω(Cl) = 35.06% ÷ 100%.

ω(Cl) = 0.3506; mass percentage of chlorine.

If we take 100 grams of the compound:

m(Cl) = ω(Cl) · m(compound).

ω(Cl) = 0.3506 · 100 g.

ω(Cl) = 35.06 g.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 35.06 g ÷ 35.45 g/mol.

n(Cl) = 0.99 mol; amount of substance.

In molecule MCl₃: n(M) : n(Cl) = 1 : 3.

n(M) = 0.33 mol; amount of unknown metal.

M(M) = m(M) ÷ n(M).

M(M) = (100 g - 35.06 g) ÷ 0.33 mol.

M(M) = 196.8 g/mol; molar mass of the gold.

4 0

3 years ago

Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass. determine the empirical formula for a comp

nexus9112 [7]

<span>Mass of nitrogen = 14.0067
</span> Mass of oxygen = 15.9994
In this compound nitrogen = 36.86 / 14.0067 = 2.63
<span>And oxygen = 63.14 / 15.9994 = 3.95 <span>
now we have: N----- 2.63 and O----3.95
by dividing both with the smallest number we get
</span></span> <span>N-------2.63/2.63 = 1<span>
<span>O-------3.95/2.63 = 1.5
To get whole numbers we multiply both by 2
</span></span></span> N= 1 x 2 = 2 And O = 1.5 x 2= 3
<span>So, the empirical formula is N</span>₂O₃.

5 0

3 years ago

The measured voltage of an electrochemical cell consisting of pure nickel immersed in a solution of Ni2+ ions of unknown concent

Verizon [17]

Answer:

[Ni²⁺] = 1.33 M

Explanation:

To do this, we need to use the Nernst equation which (in standard conditions of temperature)

E = E° - RT/nF lnQ

However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:

E = E° - 0.05916/n logQ

As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:

Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)

With this, Q:

Q = [Cd²⁺] / [Ni²⁺]

Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:

Cd(s) ------------> Cd²⁺ + 2e⁻ E°₁ = 0.40 V

Ni²⁺ + 2e⁻ -------------> Ni(s) E°₂ = -0.25 V

E° = E°₁ + E°₂

E° = 0.40 - 0.25 = 0.15 V

Now that we have all the data, we can solve for the [Ni²⁺]:

0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])

0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])

-0.017 = -0.02958 log(5/[Ni²⁺])

-0.017/-0.02958 = log(5/[Ni²⁺])

0.5747 = log(5/[Ni²⁺])

10^(0.5747) = 5/[Ni²⁺]

[Ni²⁺] = 5/3.7558

7 0

3 years ago