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3 years ago

Which of the following substances has the highest melting point?

Chemistry

1 answer:

Aleksandr [31]3 years ago

5 0

Nonane (b) has the highest melting point.

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A caveat: I'm assuming that we're dealing with the straight-chain isomers of these alkanes (specifically pentane and nonane). The straight-chain isomer of pentane (n-pentane, CH3-[CH2]3-CH3) has a melting point of -129.8 °C; the straight-chain isomer of nonane (n-nonane, CH3-[CH2]7-CH3) has a melting point of -53.5 °C. The pattern holds as you go down (or up): The more carbon atoms, the higher the melting point. So, in decreasing order of melting points here, you'd have the following: nonane > pentane > butane > ethane.

However, one structural isomer of pentane, neopentane, has a melting point of -16.4 °C, which is higher that the melting point of n-nonane despite neopentane having the same molecular formula as its straight-chain isomer. Of course, you're not to blame for coming up with this question; this is just some extra info to keep in mind.

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What is the temperature of 0.47 mol of gas at a pressure of 1.5 atm and a volume of 10.5 l ?

Whitepunk [10]

The temp is 0.002448 of the equation

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3 years ago

If the caffeine concentration in a particular brand of soda is 1.87 mg/oz, drinking how many cans of soda would be lethal? Assum

Drupady [299]

2.77mg caffeine / 1oz12oz / 1canLethal dose: 10.0g caffeine = 10,000mg caffeine First, find how much caffeine is in one can of soda, then divide that amount by the lethal dose to find the number of cans. (2.77mg caffeine / 1oz) * (12oz / 1can) = 33.24mg caffeine / 1can. (10,000mg caffeine) * (1can / 33.24mg caffeine) = 300.84 cans. Since we can't buy parts of a can of soda, then we have to round up to 301 cans. Notice how all the values were set up as ratios and how the units cancelled.

4 0

3 years ago

When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

yKpoI14uk [10]

Answer:

For a: The number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b: The mass of nitric acid formed is 54.81 grams

For c: The mass of nitric acid formed is 206 grams

Explanation:

The given chemical reaction follows:

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)$

For a:

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with $\frac{3}{1}\times 0.250=0.75mol$ of nitrogen dioxide

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.75 moles of nitrogen dioxide will contain $0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of molecules

Hence, the number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = $\frac{2}{3}\times 1.304=0.870$ moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

$0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g$

Hence, the mass of nitric acid formed is 54.81 grams

For c:

For nitrogen dioxide:

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol$

For water:

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = $\frac{1}{3}\times 4.90=1.63mol$ of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce $\frac{2}{3}\times 4.90=3.27mol$ of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

$3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g$

Hence, the mass of nitric acid formed is 206 grams

7 0

3 years ago

Elements in the same period have the same…

ch4aika [34]

Answer:

Elements in the same period have the same (number of electron shells)

8 0

2 years ago

How many grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C?

jolli1 [7]

Answer:

22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law:

P * V = n * R * T

where R is the molar constant of the gases and n the number of moles.

In this case you know:

P= 7.5 atm
V= 2.3 L
n= ?

R= 0.082 $\frac{atm*L}{mol*K}$

T= 24 °C= 297 °K (being 0°C=273°K)

Replacing:

$7.5 atm* 2.3 L=n*0.082 \frac{atm*L}{mol*K} *297K$

Solving:

$n=\frac{7.5 atm* 2.3 L}{0.082 \frac{atm*L}{mol*K} *297K}$

n=0.708 moles

Knowing that oxygen gas is a diatomic gas of molecular form O₂ and its mass is 32 g / mole, you can apply the following rule of three: if 1 mole contains 32 grams, 0.708 moles, how much mass will it have?

$mass=\frac{0.708 moles*32 grams}{1mole}$

mass= 22.656 grams

22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C

8 0

3 years ago

Read 2 more answers