What must happen to nitrogen before it can be used

How many cubic miles are 8.48E+08 gallons of water? The density of water at ambient conditions is 1.000 g/mL.

Inessa05 [86]

The volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .

<h3>Weight of one gallon of water</h3>

The weight of 1 gal of water is given as 3785 g

Mass of 8.48 x 10⁸ gal = 3785 x 8.48 x 10⁸ = 3.2 x 10¹² g

<h3>Volume of the water in cubic meters</h3>

Volume = mass/density

Volume = 3.2 x 10¹² g/1 gmL

Volume = 3.2 x 10¹² mL x 10⁻⁶ m³/mL = 3.2 x 10⁶ m³

Thus, the volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .

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7 0

2 years ago

How much (Q) heat is needed to melt 35 g of iodine? Hf = 61.7 J/g.

aev [14]

Taking into account the definition of calorimetry and latent heat, a heat of 2159.5 J is needed to melt 35 g of iodine.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

<h3>Heat needed to melt iodine</h3>

In this case, you know:

m= 35 g
L=61.7 $\frac{J}{g}$

Replacing in the definition of latent heat:

Q= 35 g× 61.7 $\frac{J}{g}$

Solving:

Finally, a heat of 2159.5 J is needed to melt 35 g of iodine.

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8 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

3 years ago

Which of the following statements is FALSE?

Alborosie

Answer:

The answer to your question is: letter E

Explanation:

A. This option is correct, the n = 3 shell only has subshells: s, p and d, and shell n = 4 or 5 have f subshell.

B. This option is true in subshell p could be at most 6 electrons and 3 suborbitals.

C. This option is correct orbital "s" is a sphere.

D. This option is correct, in subshell d could be at most 10 electrons and 5 orbitals.

E. This option is false, hydrogen only has 1 electron and then one subshell (s).

6 0

3 years ago

PLEASE HELP ME!!

blagie [28]

1, When temperature is increased the volume will also increase. this is because the particles will gain kinetic energy and bombard the walls of the container of the gas at a higher frequency, therefore, for the pressure to remain constant as per Charles' law, the volume will have to increase so that the rate of bombardment remains constant. This is explained by the Charles law which states that the volume of a gas is directly proportional to the absolute temperature provided pressure remains constant.

2. When temperature is Decreased the volume will also Decrease. this is because the particles will loose kinetic energy and bombard the walls of the container of the gas less frequently, therefore, for the pressure to remain constant as per Charles' law, the volume will have to reduce so that the rate of bombardment remains constant. This is explained by the Charles law which states that the volume of a gas is directly proportional to the absolute temperature provided pressure remains constant.

3. When temperature is increased the pressure will increase. This is because the gas particles gain kinetic energy and bombard the walls of the container more frequently. this is according to Pressure law which states that for a constant volume of a gas the pressure is directly proportional to absolute temperature

4. When temperature is decreased, pressure will decrease, This is because the gas particles lose kinetic energy and bombard the walls of the container less frequently. this is according to Pressure law which states that for a constant volume of a gas the pressure is directly proportional to absolute temperature

5. When particles are added, pressure will increase. This is because the bombardment per unit area also increases. Boyles law explains this, that at fixed temperature the volume of a gas is inversely proportional to the pressure.

6. When particles are removed, the pressure will decrease. This is because the bombardment per unit area also decreases. Boyle's law explains this, that at fixed temperature the volume of a gas is inversely proportional to the pressure.

7 0

3 years ago