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
Shkiper50 [21]
2 years ago
5

Which objects are scratched?

Chemistry
2 answers:
Solnce55 [7]2 years ago
4 0

Answer:

What objects

Explanation:

masha68 [24]2 years ago
4 0

Answer:

We would need a picture to help you out with this.

Explanation:

You might be interested in
Identify the type of molecule shown in the picture
BlackZzzverrR [31]
1) acid
2) ether
3) ester
4) aldehyde 
5) ketone
6) amine 
7) alcohol 
7 0
3 years ago
Methods scientist use to determine a change in physical or chemical change in an Item
forsale [732]
They'd observe any new substances or new color, and temp. that is irreversible. 

If this helped and you have any other questions, feel free to comment on my wall. Hope this helped.
8 0
3 years ago
At 9 a.m. on May 21, a body is discovered near a path in a park. The body is 10 feet from a tree beside the path. The body is pa
katrin2010 [14]

I'm pretty sure its D, Time of discovery of the body. Let me know (:

4 0
3 years ago
Read 2 more answers
If 0.2 g of nitrobenzene are added to 10.9 g of naphthalene, calculate the molality of the solution. (given: molar mass of nitro
In-s [12.5K]

Molality is defined as 1 mole of a solute in 1 kg of solvent.  

Molality=

\frac{Number of moles of solute}{Mass of solvent in kg}

Number of moles of solute, n=  

\frac{Given mass of the substance}{Molar mass of the substance}

Given mass of the nitrobenzene=0.2 g

Molar mass of the substance= 123.06 g mol⁻¹

Number of moles of nitrobenzene,  

n= \frac{0.2 }{123.06}

Number of moles of nitrobenzene, n= 0.0016  mol

Mass of 10.9 g of naphthalene in kg=0.0109  

Molality= \frac{0.0016}{0.0109 }

Molality= 0.146 m

7 0
3 years ago
You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar
torisob [31]

Based on Beer-Lambert's Law,

A = εcl ------(1)

where A = absorbance

ε = molar absorptivity

c = concentration

l = path length

Step 1: Calculate the concentration of the diluted Fe3+ standard

Use:

V1M1 = V2M2

M2 = V1M1/V2 = 10 ml*6.35*10⁻⁴M/55 ml = 1.154*10⁻⁴ M

Step 2 : Calculate the concentration of the sample solution

Based on equation (1) we have:

A(Fe3+) = ε(1.154*10⁻⁴)(1)

A(sample) = ε(C)(4.4)

It is given that the absorbances match under the given path length conditions, i.e.

ε(1.154*10⁻⁴)(1) = ε(C)(4.4)

C = 0.262*10⁻⁴ M

This is the concentration of Fe3+ in 100 ml of well water sample

Step 3: Calculate the concentration of Fe3+ in the original sample

Use V1M1 = V2M2

M1 = V2M2/V1 = 100 ml * 0.262*10⁻⁴ M/35 ml = 7.49*10⁻⁵M

Ans: Concentration of F3+ in the well water sample is 7.49*10⁻⁵M


7 0
3 years ago
Other questions:
  • HHHHHEEEELLLLLPPP PPPLLEEASEEE I GIVE BRAINLIEST!!!!!!!!!
    5·1 answer
  • Rusting Out. How Acids Affect the Rate of corrosion
    9·1 answer
  • what is the balanced equation when copper metal is placed in a solution when platnium ii chloride is placed. what is the equatio
    15·1 answer
  • What mass of water can be obtained from 4.0 g of H2 and 16 g of O2?2 H2 + O2 ---> 2 H2O18 g36 g54 g9 g
    11·2 answers
  • What is the HCl concentration if 52.0 mL of 0.350 M NaOH is required to titrate a 30.0 mL sample of the acid?
    7·1 answer
  • How many sulfur atoms are there in 4.45 mol of sulfur?
    8·1 answer
  • As stated in the article, “As Sticky as a Gecko . . . but Ten Times Stronger!,” the adhesive the researchers developed sticks be
    9·2 answers
  • How are reactants different than products?<br>​
    7·1 answer
  • M[oknjjom[pmnj <br> word to my mother y'all look
    12·1 answer
  • If 22.5L of nitrogen at 0.98 atm is compressed to 0.95 atm,what is the new volume?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!