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
Setler79 [48]
3 years ago
12

How are catapults an example of a third class lever? Please explain!

Chemistry
1 answer:
Katen [24]3 years ago
3 0
I hope that you can understand this!!! Lol

The thing that you have to pull back to release with, that would be considered a third class lever.

I hope this helps. :)
You might be interested in
Please help D: anyone who helps with all the chemistry problems on my page will get Brainliest Answer on all!!! I have an hour t
Pavel [41]
There are 6.022×1023 molecules in one mole of glucose so you would just divide this number by 100 and multiply the result by 54 which would leave you with 3326.67324 molecules in 0.54 moles of glucose
7 0
2 years ago
C3H7-C(=O)-NH2 IUPAC NAME ?
Yuri [45]

Answer:

Amide

Explanation:

O=NH2 is the Amide group versus NH2, which is the amine group.

7 0
3 years ago
Read 2 more answers
33) Which is the correct name for the molecule depicted below?
Morgarella [4.7K]

Answer:

C. 2,3,3,4-tetramethylhexane

Explanation:

7 0
3 years ago
You are asked to bring the pH of 0.500L of 0.550 M NH4Cl to 7.00. Which of the following solutions would you use: 12.0 M HCl or
ch4aika [34]
<span>1 drop is approximately 0.05mL. Since 0.500L of 0.550M NH4Cl contains 0.275mol of substance (calculated by using c=n/V formula), equal amount of substance of NH3 is needed to neutralize this solution (since pH of 7 is neutral solution). Thus, we need 0.0275L of NH3, i.e. around 550 drops.</span>
6 0
3 years ago
A 21.82 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21.33 grams of CO2 and 4
morpeh [17]

<u>Answer:</u> The molecular formula for the given organic compound is C_2H_2O_4

<u>Explanation:</u>

  • The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=21.33g

Mass of H_2O=4.366g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.33 g of carbon dioxide, \frac{12}{44}\times 21.33=5.82g of carbon will be contained.

  • For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.366 g of water, \frac{2}{18}\times 4.366=0.485g of hydrogen will be contained.

  • Mass of oxygen in the compound = (21.82) - (5.82 + 0.485) = 15.515 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5.82g}{12g/mole}=0.485moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = \frac{0.485}{0.485}=1

For Hydrogen  = \frac{0.485}{0.485}=1

For Oxygen  = \frac{0.969}{0.485}=1.99\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is C_1H_{1}O_2=CHO_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

  • The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Thus, the molecular formula for the given organic compound is C_2H_2O_4

4 0
3 years ago
Other questions:
  • Use the periodic table to identify the number of core electrons and the number of valence electrons in each case below. Potassiu
    6·2 answers
  • HELP! LOOK AT QUESTION!!
    8·2 answers
  • You heat a piece of iron from 200 to 400k what happens to the atoms energy of random motion
    7·1 answer
  • A student fills her burette with NaOH to the 2.5 mL mark. She titrated her sample of the NaOH until she reaches the endpoint (i.
    6·1 answer
  • An example of a substance that is acidic is __________
    5·1 answer
  • What occurs in endothermic reactions?
    14·1 answer
  • (Please someone help me it’s urgent) I need to pass this
    11·1 answer
  • How many grams of molybdenum(Mo) are in 2.68E24 atoms of Mo?
    8·1 answer
  • 6. What is the molarity of 175 mL of solution containing 2.18 grams of NazS04-10H2O?​
    10·1 answer
  • Kinetic energy is energy an object has because of its:
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!