1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
coldgirl [10]
3 years ago
12

The limiting reactants appeared to be _______ Because___________

Chemistry
1 answer:
Igoryamba3 years ago
8 0

Answer:

The limiting reagents seemed to be those that were consumed first .

Because when this reagent is consumed, the reaction stops. The quantity of this determines the total quantity of the product formed.

Explanation:

A limiting reagent is one that is consumed in its entirety. In this way, it delimits the amount of product that can be formed.

Take the case of two substances that interact and produce a chemical reaction. If one of the substances runs out as it is consumed during the process, the reaction will stop. The reagent consumed acts as a limiting reagent, that is, it limits the possibility of the reaction proceeding, and therefore it also limits the amount of the product generated by the reaction.

You might be interested in
39 which statement about the process of dissolving is TRUE
xz_007 [3.2K]
C !!!!!!!!!!!!!!!!!!!!!!
6 0
3 years ago
How many moles are present in
olasank [31]

Answer: (a) There are 0.428 moles present in 12 g of N_{2} molecule.

(b) There are 2 moles present in 12.044 \times 10^{23} particles of oxygen.

Explanation:

(a). The mass of nitrogen molecule is given as 12 g.

As the molar mass of N_{2} is 28 g/mol so its number of moles are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{12 g}{28 g/mol}\\= 0.428 mol

So, there are 0.428 moles present in 12 g of N_{2} molecule.

(b). According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

Therefore, moles present in 12.044 \times 10^{23} particles are calculated as follows.

Moles = \frac{12.044 \times 10^{23}}{6.022 \times 10^{23}}\\= 2 mol

So, there are 2 moles present in 12.044 \times 10^{23} particles of oxygen.

4 0
3 years ago
A piece of copper is 2x2x4 cm and has a mass of 142, what is the density of copper
ohaa [14]

Answer:

Demisty = mass/volume = 142/ (2×2×4) = 142 / 16 =

8.875g/cm^3

Explanation:

3 0
3 years ago
1. Write the equilibrium constant expression for the following:
melisa1 [442]

Answer:

c

Explanation:

it could honestly be wrong but I'm not sure

3 0
3 years ago
August kekule described the various ring structures of the carbon compound benzene. What type of chemist would he be considered
Andreyy89
B.  He would be considered an Organic Chemist since Organic Chemistry is the study of Carbon and its compounds.
3 0
3 years ago
Read 2 more answers
Other questions:
  • Melamine (C3N3(NH2)3) is a component of many adhesives and resins and is manufactured in a two-step
    11·1 answer
  • If 4.5 mol of C6H8O7 react, how many moles of CO2 and Na3C6H5O7 will be produced?
    15·1 answer
  • PLEASE ANSWER FAST!!!!! Consider this hypothesis statement. "If the number of hours a student studies increases, then the studen
    9·1 answer
  • What is a substance that decreases the rate of a chemical reaction called?
    14·2 answers
  • A flexible container at an initial volume of 6.13 L contains 6.51 mol of gas. More gas is then added to the container until it r
    5·1 answer
  • Which molecule, when added to a solution, would give the solution the highest pH
    13·1 answer
  • An ionic compound whose aqueous solution conducts an electric current is called a(n) ______________.
    6·2 answers
  • please someone describe the graph of the cooling curve of water !!!!!!!!!!!!!!!! Please answer !!!!!!!!!! HELP
    11·1 answer
  • PLEASE HELP SCIENCE SCIENCE HELP PLEASE PLEASE HELP SCIENCE SCIENCE HELP PLEASE PLEASE HELP SCIENCE SCIENCE HELP PLEASE PLEASE H
    11·2 answers
  • Class material don't interact please
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!