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3 years ago

What are the products when ammonia (NH3) decomposes according to this equation? NH3

1 answer:

7 0

Ammonia (NH3) is the combination of Nitrogen and Hydrogen elements.
=> N2 + 3H2 => 2NH3

Ammonia is basically used as a fertilizer. It is a gas composed of nitrogen and hydrogen. It is colorless with strong odor. Here are some other uses of Ammonia aside from fertilizer:
=> used by manufacturer to produce synthetic fiber
=> Used in metallurgical process
Ammonia can be decomposed easily and it produce hydrogen that is very convenient in welding.
Ammonia’s boiling point is -28.03 F and freezing point is -107.8F.

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Which type of molecule is shown below?

romanna [79]

Answer: C

Explanation:

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3 years ago

Read 2 more answers

When elemental sodium reacts with water, sodium hydroxide and hydrogen will form. What mass of sodium (in grams) must be reacted

tangare [24]

Answer:

0.703g Na must reacted

Explanation:

The reaction of Sodium, Na, With water, H₂O is:

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

Where 2 moles of sodium reacts with an excess of water to produce 1 mole of hydrogen

To solve this question we have to use PV = nTR to solve the moles of the gas. With the moles of hydrogen we can find the moles of sodium that reacted and its mass:

Moles H₂:

PV = nRT

PV /RT = n

Where P is pressure = 1.15atm

V is volume in liters = 0.325L

R is gas constant = 0.082atmL/molK

T is absolute temperature = 25°C + 273 = 298.0K

1.15atm*0.325L / 0.082atmL/molK*298.0K = n

0.0153 moles of hydrogen are produced

Moles Na:

0.0153 moles H₂ * (2moles Na / 1mol H₂) = 0.0306 moles Na

Mass Na -Molar mass: 22.99g/mol-:

0.0306 moles Na * (22.99g / mol) =

<h3>0.703g Na must reacted</h3>

8 0

3 years ago

Starting with 9.3 moles of O2, how many moles of H2S will be needed and how many moles of SO2 will be produced in the following

tensa zangetsu [6.8K]

Answer: The amount of hydrogen sulfide needed is 6.2 moles and amount of sulfur dioxide gas produced is 6.2 moles

Explanation:

We are given:

Moles of oxygen gas = 9.3 moles

The chemical equation for the reaction of oxygen gas and hydrogen sulfide follows:

$2H_2S+3O_2\rightarrow 2SO_2+2H_2O$

For hydrogen sulfide:

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 2 moles of hydrogen sulfide

So, 9.3 moles of oxygen gas will react with = $\frac{2}{3}\times 9.3=6.2mol$ of hydrogen sulfide

For sulfur dioxide:

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 2 moles of sulfur dioxide

So, 9.3 moles of oxygen gas will produce = $\frac{2}{3}\times 9.3=6.2mol$ of sulfur dioxide

Hence, the amount of hydrogen sulfide needed is 6.2 moles and amount of sulfur dioxide gas produced is 6.2 moles

7 0

3 years ago

Calculate the mass of aluminum bromide that can be produced from 4.4 mil of bromide

Aloiza [94]

Answer:

197 grams

Explanation:

Without taking into account the number of sig figs you have for your values, you can say that the answer will once again be

197 g of aluminium bromide.

(I hope this is right)

7 0

3 years ago

2 questions. will give brainliest if i can figure out how

sp2606 [1]

1= b. electrons are lost

2= a. after one substance is oxidized

5 0

3 years ago