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3 years ago

Which topic is Eunice's project most likely about?

Chemistry

1 answer:

otez555 [7]3 years ago

6 0

Answer:

pioneer species. because they are flower like species

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Which of the statements regarding alcohols and ethers are true? 1-Propanol is more likely to be a liquid at 50 °C50 °C than is m

irina [24]

Answer:

Following statements are true:

1-Propanol is more likely to be a liquid at 50 than is methoxyethane (ethyl methyl ether).

Alcohol molecules can form hydrogen bonds among themselves;ether molecules cannot

Explanation:

1-Propanol is more likely to be a liquid at 50 than is methoxyethane (ethyl methyl ether).

This is true

Propanol has a boiling temperature (this is the temperature at which the vapor pressure of a liquid equals the pressure surrounding the liquid and the liquid changes into a vapor) of 97-98 °C while methoxyethane a boiling point of 7.4 °C.

This means at 50 °C propanol is liquid but methoxyethane at 50°C is a gas.

Methoxyethane (ethyl methyl ether) is more soluble in water than is 1-propanol.

This false:

Propanol has a hydrogen bonding. The OH on the alcohol can form a hydrogen bond (as a hydrogen acceptor and donor). Water can hydrogen bond. This means alcohol is will disolve in water. ( Remember Like dissolves like). So, when you have a molecule that can act as a donor and acceptor you end up with a much better interaction with water and end up with something that is more miscible. Ethers are only acceptors since they do not have a hydrogen attached to the oxygen. So, they can only accept a proton from water.

Methoxyethane (ethyl methyl ether) is more likely to be a liquid at 50 °C50 °C than is 1-propanol

This is false because propanol has a boiling temperature (this is the temperature at which the vapor pressure of a liquid equals the pressure surrounding the liquid and the liquid changes into a vapor) of 97-98 °C while methoxyethane a boiling point of 7.4 °C.

This means at 50 °C propanol is liquid but methoxyethane at 50°C is a gas.

Alcohol molecules can form hydrogen bonds among themselves;ether molecules cannot

This is true because:

Ether molecules cannot form hydrogen bonding between themselves because the oxygen atom is attached to 2 alkyl or aryl groups R-O-R, e.g diethyl ether, or Ar-O-Ar, such as diphenylether (C6H5)2O

C2H5-O-C2H5,m or dimethyl ether CH3—O-CH3. however they can between ether molecules and other molecules containing hydrogen atoms, such as alcohols, CH3OH. Any molecule with the hydrogen attached to a strongly electronegative element e.g Nitrogen or fluorine.

Methoxyethane (ethyl methyl ether) has a higher boiling point than 1-propanol.

This is false: Methoxyethane has a boiling point of 7.4 °C while propanol has a boiling point of 97-98 °C. This can be shown because at roomtemperature Methoxyethane is a gas but propanol is a liquid.

6 0

2 years ago

Question 1(Multiple Choice Worth 4 points)

zubka84 [21]

Answer

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of $H_{2}O$ over Nine moles of $O_{2}$

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Answer 4 : Mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

Solution

Solution 1 : Given,

Given mass of $Fe_{2}O_{3}$ = 55 g

Molar mass of $Fe_{2}O_{3}$ = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of $Fe_{2}O_{3}$ = $\frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}}$ = $\frac{55 g}{159.69 g/mole}$ = 0.344 moles

Balanced chemical reaction is,

$Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)$

From the given reaction, we conclude that

1 mole of $Fe_{2}O_{3}$ gives → 3 moles of CO

0.344 moles of $Fe_{2}O_{3}$ gives → 3 × 0.344 moles of CO

= 1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

= 1.032 × 28.01

= 28.90 g

Solution 2 : The balanced chemical reaction is,

$2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O$

From the given reaction, we conclude that the Six moles of $H_{2}O$ over Nine moles of $O_{2}$ is the correct option.

Solution 3 : The balanced chemical reaction is,

$4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}$

From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Solution 4 : Given,

Given mass of $Zn(ClO_{3})_{2}$ = 150 g

Molar mass of $Zn(ClO_{3})_{2}$ = 232.29 g/mole

Molar mass of $O_{2}$ = 31.998 g/mole

Moles of $Zn(ClO_{3})_{2}$ = $\frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}}$ = $(\frac{150\times 1}{232.29})moles$

The balanced chemical equation is,

$Zn(ClO_{3})_{2}}\rightarrow ZnCl_{2}+3O_{2}$

From the given balanced equation, we conclude that

1 mole of $Zn(ClO_{3})_{2}$ gives → 3 moles of $O_{2}$

$(\frac{150\times 1}{232.29})moles$ of $Zn(ClO_{3})_{2}$ gives → $[(\frac{150\times 1}{232.29})\times 3] moles$ of $O_{2}$

Mass of $O_{2}$ = Number of moles of $O_{2}$ × Molar mass of $O_{2}$ = $[(\frac{150\times 1}{232.29})\times 3] \times 31.998 grams$

Therefore, the mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Solution 5 : Given,

Number of moles of $Na_{2}SO_{4}$ = 4.2 moles

Balanced chemical equation is,

$H_{2}SO_{4}+2NaCN\rightarrow 2HCN+Na_{2}SO_{4}$

From the given chemical reaction, we conclude that

1 mole of $Na_{2}SO_{4}$ obtained from 2 moles of NaCN

4.2 moles of $Na_{2}SO_{4}$ obtained → 2 × 4.2 moles of NaCN

Therefore,

The moles of NaCN needed = 2 × 4.2 = 8.4 moles

3 0

3 years ago

Read 2 more answers

When one atom contributes both bonding electrons in a single covalent bond, the bond is called a(n) ____?

Goshia [24]

I belive this is known as coordinate covalent bond.
hope this helps!

6 0

3 years ago

What makes the smell in chlorine?

padilas [110]

Pool smell is due, not to chlorine, but to chloramines, chemical compounds that build up in pool water when it is improperly treated. Chloramines result from the combination of two ingredients: (a) chlorine disinfectants and (b) perspiration, oils and urine that enter pools on the bodies of swimmers.

7 0

3 years ago

At the Henry's Law constant for carbon dioxide gas in water is . Calculate the mass in grams of gas that can be dissolved in of

Dvinal [7]

The question is incomplete, here is the complete question:

At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.

Answer: The mass of carbon dioxide that can be dissolved is 1.7 grams

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$