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nekit [7.7K]
3 years ago
5

Fav music Artist???????????

Chemistry
2 answers:
alexgriva [62]3 years ago
6 0
Nct and sza ………………..
vladimir1956 [14]3 years ago
5 0

uhhh i’m not sure

Explanation:

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What methods do scientists use to determine if a change was physical or chemical?
Bess [88]
A physical change will not change the chemical make-up of the material. the change will not produce a new substance.

a chemical change will have a reaction that produces heat, light, a change in color, releases a gas, or rust
5 0
3 years ago
How much carbon dioxide will be formed if 12.5 grams of oxygen reacts with 7.2 grams of propane (C3H8 )? Balanced equation: C3H8
solong [7]

Answer:

10.3125 grams of carbon dioxide will be formed.

Explanation:

The balanced reaction is:

C₃H₈ + 5 O₂→ 3 CO₂ + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of the compounds:

  • C₃H₈: 44 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

then by stoichiometry of the reaction, the following amounts of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole = 44 g
  • O₂: 5 moles* 32 g/mole= 160 g
  • CO₂: 3 moles* 44 g/mole= 132 g
  • H₂O: 4 moles* 18 g/mole= 72 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: If by reaction stoichiometry 44 grams of propane react with 160 grams of oxygen, 7.2 grams of propane react with how much mass of oxygen?

mass of oxygen=\frac{7.2 grams of propane*160 grams of oxygen}{44 grams of propane}

mass of oxygen= 26.18 grams

But 26.18 moles of O₂ are not available, 12.5 grams are available. Since you have less mass than you need to react with 7.2 grams of propane, oxygen O₂ will be the limiting reagent.

Then you can apply the following rule of three: if by stoichiometry of the reaction 160 grams of oxygen form 132 grams of carbon dioxide, 12.5 grams of oxygen will form how much mass of carbon dioxide?

mass of carbon dioxide=\frac{12.5 grams of oxygen*132 grams of carbon dioxide}{160 grams of oxygen}

mass of carbon dioxide= 10.3125 grams

<u><em>10.3125 grams of carbon dioxide will be formed.</em></u>

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7 0
3 years ago
On a cool morning, Uyen’s breath can form a cloud when she breathes out. Which changes of state are most responsible for Uyen se
Dima020 [189]
Condensation because condensation is a process of freezing
8 0
3 years ago
Read 2 more answers
Nitrogen monoxide and water react to form ammonia and oxygen, like this: (g)(g)(g)(g) Write the pressure equilibrium constant ex
xxTIMURxx [149]

Answer:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

Explanation:

First, we have to write the balanced chemical equation for the reaction. Nitrogen monoxide (NO) reacts with water (H₂O) to give ammonia (NH₃) and oxygen (O₂), according to the following:

NO(g) + H₂O(g) → NH₃(g) + O₂(g)

To balance the equation, we add the stoichiometric coefficients (4 for NH₃ and NO to balance N atoms, then 6 for H₂O to balance H atoms and then 5 for O₂ to balance O atoms):

4 NO(g) + 6 H₂O(g) → 4 NH₃(g) + 5 O₂(g)

All reactants and products are in the gaseous phase, so the equilibrium constant is expressed in terms of partial pressures (P) and is denoted as Kp. The Kp is expressed as the product of the reaction products (NH₃ and O₃) raised by their stoichiometric coefficients (4 and 5, respectively) divided into the product of the reaction reagents (NO and H₂O) raised by their stoichiometric coefficients (4 and 6, respectively). So, the pressure equilibrium constant expression is written as follows:

Kp = \frac{P(NH_{3}) ^{4} P(O_{2}) ^{5}}{P(NO) ^{4} P(H_{2}O)^{6}}

6 0
3 years ago
What volume of oxygen gas is needed to completely combust 0.202L of butane (C4H10) gas?
saveliy_v [14]

Answer:

look at the file :)

Explanation:

4 0
3 years ago
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