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3 years ago

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

1 answer:

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. :)

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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]

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.

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]

Answer: The molecular formula for the given organic compound is $C_2H_2O_4$

Explanation:

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:

Step 1: 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 = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles$

Step 2: 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 = $\frac{0.969}{0.485}=1.99\approx 2$

Step 3: 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