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
Elis [28]
3 years ago
15

Can some body plz help correct me or correct the true or false answers I did wrong. I’m not sure there right :(

Chemistry
1 answer:
3241004551 [841]3 years ago
8 0

4 3 and 5 are wrong:)

4 is true and 5 is false

4. insulators are really thick materials that keep electrons within their space.

For example, a charger the rubber is the insulators and the electrons are flowing inside of the charger so you don't get shocked.

5. without the force of attraction, there are no atoms:)

for number 3, its a trick question

Hope this helped!

(learned these this year so I didn't forget them:)

You might be interested in
What is called a bond between a metal and a nonmetal
sasho [114]
They are not similar at all they are completely different. It even says it in the words
6 0
3 years ago
Read 2 more answers
a 40-lb container of peat moss measure 14x20x30 inches and has an average density of 0.13 g/cm^3. how many bags of peat moss are
telo118 [61]

Volume of the peat moss = 14\times 20\times 30 inches

= 8400 in^{3}

Convert the above volume into cm^{3}

1 in^{3}= 16.4 cm^{3}

Thus, volume in cm^{3} is:

Volume of peat moss =  8400 in^{3}\times \frac{16.4 cm^{3}}{1 in^{3}}

= 137760 cm^{3}

Now,

Total volume by using area and depth of the peat moss = area of peat moss \times depth of peat moss

= (13 ft \times  25 ft)\times 1.9 inches

= (325 ft^{2})\times 1.9 inches

Convert above values in cm to get the value of volume in cm^{3}:

1 ft= 30.48 cm

1 in= 2.54 cm

Thus, volume in cm^{3} is:

Total volume = (325 ft^{2}\times\frac{(30.48 cm)^{2}}{(1 ft)^{2}})\times (1.9 in\times \frac{2.54 cm}{1 in})

= 301934.88 cm^{2}\times 4.826 cm

= 1457137.73088 cm^{3}

Now, number of bags is calculated by the ratio of total volume of the peat moss to the volume of the peat moss.

Number of bags  =\frac{total volume of peat moss}{volume of peat moss}

Substitute the values of volume in above formula:

Number of bags  = \frac{1457137.73088 cm^{3}}{137760 cm^{3}}

= 10.57

≅ 11 bags

Thus, number of bags of peat moss are needed to cover an area measure 13 ft by 25 ft to a depth of 1.9 inches are 11 bags.


4 0
3 years ago
How are scientific questions answered? OA. Through feelings and guesses
Tanzania [10]
D. through measuring and observing !
8 0
3 years ago
Read 2 more answers
What would happenen if we were to have a solar eclipse and a lunar eclipse
stepladder [879]
We won't be able to see the moon
6 0
3 years ago
PLZ HELP ASAP
Verizon [17]

Answer:

Using glucose to provide energy for the body

Explanation:

3 0
3 years ago
Other questions:
  • Name any two waste materials of the plants​
    7·2 answers
  • How many oxygen atoms are present in 2.28 moles of nitric<br> acid, HNO3 ?
    9·1 answer
  • A scientist isolates 2.391g of a gas. The sample occupies a volume of 562.3mL at 94.8 degrees celsius and 108.2 kPa. Calculater
    15·1 answer
  • A/an _______ bond allows metals to conduct electricity
    6·2 answers
  • Complete the table for ion charge based upon their losing or gaining electrons in the outer shell. (Use the periodic table as ne
    8·1 answer
  • Pls help what the ratio ?.<br>enter your answer as an integer ​
    7·1 answer
  • Describe how an electromagnetic wave carries energy from place to place. Your description should include the terms electric fiel
    8·1 answer
  • What are automated weather stations?
    11·1 answer
  • I need help with both are they right please read the questions my grades dropped and if I don’t get them up I can’t go out of to
    13·1 answer
  • During a reaction, the enthalpy of formation of an intermediate is 34 kJ/mol. During the reaction, 4 moles of the intermediate a
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!