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
Gnoma [55]
3 years ago
12

What is the aim of golden rain demonstration​

Chemistry
1 answer:
Andrei [34K]3 years ago
3 0

demonstrates the formation of a solid precipitate

You might be interested in
Which response is a positive way to manage stress?
Xelga [282]

Answer: Eating healthy is a postive way

Explanation:

6 0
3 years ago
Identify the sample and analyte in each of the scenarios.
MrMuchimi

Answer:

a) Analyte: lead. Sample: paint.

b) Analyte: nitrate. Sample: soil.

c) Analyte: citric acid. Sample: Lime

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

Explanation:

A sample is a portion of material selected from a larger quantity of material while an analyte is the chemical of the system that will be analysed.

Thus:

a) Analyte is lead while you must take a sample of paint to analyze this lead.

b) Analyte is the nitrate while sample must be soil.

c) Analyte is citric acid and lime is the sample

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

7 0
3 years ago
How much water is in lakes, rivers and swamps?
Juli2301 [7.4K]

Answer:

That depends on the size of it

Explanation:

3 0
3 years ago
Why is it important to test sending an electronic résumé before sending it to an employer?
cluponka [151]
To ensure there are no mistakes and it is polished and well. A non-tested resume may cost you a job opportunity.
6 0
3 years ago
Read 2 more answers
Use the following reaction:
juin [17]

Answer:

78.4g

Explanation:

Given parameters:

Mass of copper (II) chloride  = 90g

Mass of sodium nitrate  = 120g

Unknown:

Mass of sodium chloride that can be formed  = ?

Solution:

The balanced chemical reaction is:

            CuCl₂   +   2NaNO₃ →   Cu(NO₃)₂   +   2NaCl

Let us find the limiting reactant. This reactant will determine the extent of the reaction of the amount of product that will be formed.

First, convert the masses to number of moles;

      Number of moles  = \frac{mass}{molar mass}

Molar mass of   CuCl₂  = 63.5 + 2(35.5) = 134.5g/mol

Molar mass of NaNO₃   = 23 + 14 + 3(16) = 85g/mol

  Number of moles of CuCl₂ = \frac{90}{134.5}  = 0.67mole

  Number of moles of NaNO₃  = \frac{120}{85}  = 1.41mole

From the balanced reaction equation:

      1 mole of CuCl₂   would react with 2 moles of NaNO₃

     0.67mole of CuCl₂   would require   0.67 x 2  = 1.34mole of NaNO₃

So, CuCl₂ is the limiting reactant

     1 mole of CuCl₂ will produce 2 mole of NaCl

    0.67mole of CuCl₂ will therefore yield 0.67 x 2 = 1.34mole of NaCl

Mass of NaCl = number of moles x molar mass

  Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

 

Mass of NaCl  = 58.5 x 1.34  = 78.4g

5 0
3 years ago
Other questions:
  • If the kitchen door is closed, how will that affect your ability to smell cooking odors in your room?
    12·2 answers
  • Which evidence best supports scientists' inferences about the origin and age of the universe?
    10·2 answers
  • Exit What is the number of moles of solute in 250 mL of a 0.4 M solution? Don't forget the unit.
    12·1 answer
  • PLEASE HELPP
    6·1 answer
  • Will a marble keep rolling forever on a flat hardwood floor?
    6·1 answer
  • How many electrons in an atom could have these sets of quantum numbers n=7 l=3 ml=-1?
    7·1 answer
  • C2H6O<br> Percent Composition with work shown
    10·1 answer
  • PLS HELP I REALLY NEED IT IM ON THE VERGE OF TEARS
    12·1 answer
  • What are the the basic cloud types?
    13·1 answer
  • How many grams of glucose are in 4.5x1023 molecules of glucose?:
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!