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2 years ago

An object has________when it has been stretched or compressed

Chemistry

1 answer:

Alina [70]2 years ago

4 0

Answer:

Elasticity or Flexibility

Explanation:

The ability of an object to resume its normal shape after being stretched or compressed is called elasticity

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A. Which line represents travel at the fastest speed? Justify your answer.__

Vikentia [17]

Answer:

Which line represents travel at the fastest speed? Justify your answer.__

b. Describe the line that should be added for an object that is not moving with respect to the chosen reference point.__

Answer the question using 5 complete sentences and the vocabulary terms; velocity, position, and distance

Which line represents travel at the fastest speed? Justify your answer.__

b. Describe the line that should be added for an object that is not moving with respect to the chosen reference point.__

Answer the question using 5 complete sentences and the vocabulary terms; velocity, position, and distance

Explanation:

5 0

2 years ago

The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

Answer: The molecular formula of the compound is $C_4H_{10}O_4$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

Calculating the molecular formula:

The chemical equation for the combustion of compound 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 $CO_2=36.26g$

Mass of $H_2O=14.85g$

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

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

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 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

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 = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

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

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

3 years ago

How is it possible for helium to have 8 spectral lines when it only has 2 electrons?

neonofarm [45]

It has more lines in it compared to hydrogen emission spectrum. It is mainly because the helium atom has more electrons than a hydrogen atom. Therefore, more electrons get excited when we pass a white light beam through a helium sample, and it causes the emission of more spectral lines

5 0

2 years ago

What is the pressure in a 24.0-LL cylinder filled with 32.7 gg of oxygen gas at a temperature of 319 KK? Express your answer to

Blizzard [7]

Answer:

The pressure in that cylinder = 1.12atm

Explanation:

We use general gas law to calculate it. General gas law is gotten by combining Boyle's law, Charles' law and Avogadro's law. Thus

P = nRT/V

Where n = number of moles

R = the gas constant

T is the Temperature, V is the volume and P is the pressure.

Given: T = 319K, V = 24L, R = 0.0821 L.atm/K.mol

The first step is to find n using

n = mass of O2/molar mass of O2

=32.7/32

=1.0219

Now, using P =nRT/V

P = 1.0219 ×0.0821×319÷24

Therefore P = 1.12atm

8 0

3 years ago

In this photograph what rock formation is presented by the blue lines?

Effectus [21]

It is representing syncline rock formation

there are two rock formation based on fold formation : syncline and anticline

In syncline rock formation there is fold like trough unlike anticline where it is like crust

In syncline the fold is downward as shown in photo and the new rock is outer fold and old at inner side

3 0

3 years ago

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