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
IgorC [24]
3 years ago
11

Should a mortar and pestle should be used for grinding only one substance at a time

Chemistry
1 answer:
MatroZZZ [7]3 years ago
8 0
The mortar and pestle is most commonly used in chemical laboratories or in the kitchen. Its key function is to grind the material into smaller pieces, usually into its powdered form. It looks like that shown in the picture. The mortar is the bowl in which the material to be pounded is placed, and the pestle does the pounding.

Now, when you ask if it can only pound one at a time, my honest answer is, it depends. Depending on the size of your mortar, you could grind materials two or three at a time. But if you are concerned with contamination, then you do it one at a time, especially if you don't want them to get mixed up.

You might be interested in
Which of the following statements is/are correct?
7nadin3 [17]

Answer:

All are correct

Explanation:

1) The angular momentum quantum number, l, are the subshells within a shell (principle quantum number) it talks about the "form" of an orbital, the number itself tells you about the number of angular nodes (a plane without electronic density). It starts at l=0 where you don't see any nodes and it takes the form of an sphere, and we knowing it bu another name an s-orbital. It takes values up to n-1.

l=0 (sphere - s-orbital)

l=1 (p-orbital)

l=2 (d-orbital)

2) The magnetic quatum number, ml relates to the number of orbitals within a subshell then it is related with l, taking values form -l to l incluing 0.

For l=0 (s-orbital) ml=0

For l=1 (p-orbital) ml=1,0,-1

For l=2 (d-orbital) ml=2,1,0,-1,-2

3) In every shell we are restricted by the total number of nodes of any orbital. Then if we want a d-orbital with l=3 we need at least 3 plane nodes only achievable with n=3 at least.

8 0
3 years ago
A catalyst is used to increase the reaction rate of a chemical reaction.
hichkok12 [17]
The answer would be C
3 0
3 years ago
If 9 moles of P203 are formed, how many<br> moles of O2 reacted?
masha68 [24]
9 x 3 = 27
27 moles of O reacted
27 / 2 = 13.5 O2 reacted
round up to 14 moles of O2
3 0
2 years ago
5. The chart lists organisms in five different categories living near the Texas
Nata [24]

Answer:

A

Explanation:

3 0
2 years ago
Read 2 more answers
a sample of 25.0g of an unknown metal is added to 25.0ml of water in a graduated cylinder and the final volume is 28.5ml what is
leva [86]

Answer:

<h2>The answer is 7.14 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass of metal = 25 g

volume = final volume of water - initial volume of water

volume = 28.5 - 25 = 3.5 mL

It's density is

density =  \frac{25}{3.5}  \\  = 7.142857...

We have the final answer as

<h3>7.14 g/mL</h3>

Hope this helps you

8 0
3 years ago
Other questions:
  • Is my answer correct?
    15·2 answers
  • Is MgCO3 a correct chemical formula
    15·1 answer
  • what type of wave uses thioglycolic acid or its derivatives with ammonia and procceses the hair without heat?
    14·1 answer
  • Identify the molecules with a dipole moment: (a) SF (b) CF (c) CCCB (d) CHCI (e) H.CO
    5·1 answer
  • Whag is the answer. please i need help​
    6·1 answer
  • "accelerated erosion, or change in vegetation cover" tend to persist and grow over time, rather than a return to the pre-disturb
    13·1 answer
  • Help me plsssss
    14·1 answer
  • Which of the following would be considered a compound?<br> He<br> CI<br> NaCl<br> B
    10·1 answer
  • Kate and Janie are about to pour water into a beaker when they notice
    11·2 answers
  • Which of the following statements is true?
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!