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2 years ago

Are all the molecules in the picture the same?

Chemistry

2 answers:

Dvinal [7]2 years ago

3 0

Answer:

There is no picture.

ioda2 years ago

3 0

Answer:

yes because they are together yp diffrent moleucoles

Explanation:

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Larisa pumps up a soccer ball until it has a gauge pressure of 61 kilopascals. The volume of the ball is 5.2 liters. The air tem

Nuetrik [128]

<h3>Answer:</h3>

B. 0.33 mol

<h3>Explanation:</h3>

We are given;

Gauge pressure, P = 61 kPa (but 1 atm = 101.325 kPa)

= 0.602 atm

Volume, V = 5.2 liters

Temperature, T = 32°C, but K = °C + 273.15

thus, T = 305.15 K

We are required to determine the number of moles of air.

We are going to use the concept of ideal gas equation.

According to the ideal gas equation, PV = nRT, where P is the pressure, V is the volume, R is the ideal gas constant, (0.082057 L.atm mol.K, n is the number of moles and T is the absolute temperature.

Therefore, to find the number of moles we replace the variables in the equation.
Note that the total ball pressure will be given by the sum of atmospheric pressure and the gauge
Therefore;
Total pressure = Atmospheric pressure + Gauge pressure

We know atmospheric pressure is 101.325 kPa or 1 atm

Total ball pressure = 1 atm + 0.602 atm

= 1.602 atm

That is;

PV = nRT

n = PV ÷ RT

therefore;

n = (1.602 atm× 5.2 L) ÷ (0.082057 × 305.15 K)

= 0.3326 moles

= 0.33 moles

Therefore, there are 0.33 moles of air in the ball.

4 0

2 years ago

The percent composition by mass of a compound is 76.0% c, 12.8% h, and 11.2% o. the molar mass of this compound is 284.5 g/mol.

Leokris [45]

You have a few steps to solve this one. First, we'll find the molar mass by percentage of each element in the molecule. Then, we'll divide each of those relative masses by the atomic mass of each element. The number of times the mass divides into the relative mass is the number of atoms of that element in the molecule:

C: 284.5 x .76 = 216.22
H: 284.5 x .128= 36.416
O: 284.5 x .112 = 31.864.

Now we divide out each element's atomic mass (from the periodic table). it's okay if they're approximated from the decimal answer.
C: 216.22 ÷ 12.011 ≈ 18
H: 36.416 ÷ 1.008 ≈36
O: 31.864 ÷ 15.999 ≈ 2

Therefore, the molecular formula is C18H36O2.

The empirical formula would be found by dividing out all factors of those subscript numbers. In our case, all of them can be divided by 2. The empirical formula would be C9H18O

7 0

2 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

<u>Answer:</u> The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

<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 $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 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 = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

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

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

<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 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

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

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

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 = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

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

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

2 years ago

What volume of 0.500 M HNO₃(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?

USPshnik [31]

Answer:

v = 500 milliliters

Explanation:

$HNO_{3}$ ⇒ $H^{+} + NO^{3-}$

$KOH$ ⇒ $K^{+} + OH^{-}$

1 $H^{+}$ to 1 $OH^{-}$

$\frac{0,5}{V} = \frac{0,1}{100} \\\\v * 0,1 = 50\\v = 500 milliliters$

4 0

2 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

2 years ago