Answer:
The Ideal Gas Law cannot be applied to liquids. The Ideal Gas Law is #PV = nRT#. That implies that #V# is a variable. But we know that a liquid has a constant volume, so the Ideal <u><em>Gas Law cannot apply to a liquid.</em></u>
Explanation:
this is my awnser soory if it was a multiple choice question plz mark brainliest
Answer:
the answer is true
Explanation:
it is the smallest particle in an element that takes part in a chemical reaction
Answer: A balanced equation for the given reaction is
.
Explanation:
The reaction equation will be as follows.

Number of atoms on the reactant side is as follows.
Number of atoms on the product side is as follows.
Since number of atoms on both the reactant and product sides are equal. Hence, the reaction equation is balanced.
Thus, we can conclude that a balanced equation for the given reaction is
.
The Oxyfuel gas or flame refers to a group of welding processes that use the flame produced by the combination of a fuel gas and oxygen as the source of heat.
<u>Explanation:</u>
- Oxy-fuel welding is a process that utilizes fuel gases and oxygen to weld metals. Oxyfuel gas or flame refers to a group of welding processes that utilize the flame delivered by the blending of fuel gas and oxygen as the source of heat.
- This flame is utilized for cutting and welding of two metallic pieces. This is done due to the heat produced by cutting and welding of two metallic pieces together by heating to the melting point.
- An oxyhydrogen flame is utilized for cutting and welding of two metallic pieces due to the heat produced by the flame, i.e, 2800 ° C. At this temperature, the metal gets softened effectively and thus it can easily separate or welded together.
Answer:
17.65 grams of O2 are needed for a complete reaction.
Explanation:
You know the reaction:
4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O
First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:
- N: 14 g/mol
- H: 1 g/mol
- O: 16 g/mol
So, the molar mass of the compounds in the reaction is:
- NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
- O₂: 2*16 g/mol= 32 g/mol
- NO: 14 g/mol + 16 g/mol= 30 g/mol
- H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol
By stoichiometry, they react and occur in moles:
- NH₃: 4 moles
- O₂: 5 moles
- NO: 4 moles
- H₂O: 6 moles
Then in mass, by stoichiomatry they react and occur:
- NH₃: 4 moles*17 g/mol= 68 g
- O₂: 5 moles*32 g/mol= 160 g
- NO: 4 moles*30 g/mol= 120 g
- H₂O: 6 moles*18 g/mol= 108 g
Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

mass of O₂≅17.65 g
<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>