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2 years ago

13

student has two balloons, one filled with helium and one filled with argon. Each balloon has a volume of 22.4-L at STP. Which of

the following correctly describes these samples

1 answer:

slamgirl [31]2 years ago

4 0

The statement that can be made about the samples is that each balloon contains 1 mole of each gas.

<h3>What is molar volume?</h3>

Molar volume refers to the volume that is occupied by one mole of a gas. According to Avogadro, the volume occupied by one ole of a gas is 22.4-L at STP.

Hence, if 22.4-L at STP is the volume of both the helium and argon filled balloons, then each balloon contains one mole of the gas.

Learn more about molar volume: brainly.com/question/4172228

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Iron has a density of 7.86 g/cm3 (1 cm3=1 mL). Calculate the volume (in dL) of a piece of iron having a mass of 4.54 kg . Note t

lilavasa [31]

<span>The volume of iron is obtained from the density formula (density= mass/ volume) given the density and mass. In this case, 4,540 grams of iron and a density of 7.86 g/ml are given. The volume obtained from formula is 493.64 ml or 5.78 dL. </span>

8 0

2 years ago

Who started water conservation in himachal pradesh

oee [108]

Explanation:

There are government khatris as well, which are maintained by the panchayat. Kuhls are a traditional irrigation system in Himachal Pradesh- surface channels diverting water from natural flowing streams (khuds). A typical community kuhl services six to 30 farmers, irrigating an area of about 20 ha.

3 0

3 years ago

A sample of gas occupies a volume of 4.5 L at 20 Celsius. What volume will it occupy at -15 celsius, if the pressure remains con

nataly862011 [7]

Answer: 3.96L

Explanation:

PV = nRT

The way I like to do it, it's I get rid of whatever I do not need.

We have to do an equation for sure to make both sides equal and find our desired result.

The pressure is constant, so it would be constant in both sides, therefore, there is no point in using it.

The R is a constant used in both sides so there is no point on using it either.

We don't need to work with moles in this case, so let's forget about the moles.

Therefore we are left with only V (volume ) and T(Temperature)

Which would logically make this:

V1/T1 = V2/T2

OR

T1/V1 = T2/V2

Both of these would work, we always use whatever makes our calculations easier without complicating our lives with the algebra.

That should remind you of Charles' Law

Transform degrees into kelvins

20 + 273.15 = 293.15K

-15 + 273.15 = 258.15K

So, 4.5L/293.15K = V2/258.15K

V2 = (4.5L/293.15K) x 258.15K

V2 = 3.96L

8 0

2 years ago

Calculate the wavelength for the transition from n = 4 to n = 2, and state the name given to the spectroscopic series to which t

finlep [7]

Answer:

The wavelength for the transition from n = 4 to n = 2 is<u> 486nm</u> and the name name given to the spectroscopic series belongs to <u>The Balmer series.</u>

Explanation

lets calculate -

Rydberg equation- $\frac{1}{\pi } =R(\frac{1}{n_1^2} -\frac{1}{n_2^2})$

where , $\pi$ is wavelength , R is Rydberg constant ( $1.097\times10^7$ ), $n_1$ and $n_2$ are the quantum numbers of the energy levels. (where $n_1=2 , n_2=4$ )

Now putting the given values in the equation,

$\frac{1}{\pi }=1.097\times10^7\times(\frac{1}{2^2} -\frac{1}{4^2} )$ $=2056875m^-^1$

Wavelength $\pi =\frac{1}{2056875}$

= $4.86\times10^-^7$ = 486nm

<u> Therefore , the wavelength is 486nm and it belongs to The Balmer series.</u>

8 0

3 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

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 $CO_2=3.095g$

Mass of $H_2O=1.902g$

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 $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

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 $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

2 years ago