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3 years ago

How are the IUPAC molecular naming system and the zoological taxonomic naming system similar and how are they different?

2 answers:

8 0

Well the IUPAC naming system is for naming organic compounds in chemistry whereas the taxonomic hierarchical naming system is for classifying and naming species of organisms and finding relationships in terms of physical characteristics. The only way I'd say that they're similar is in the way they actually name the compound/species. E.g, in chemistry, you always have a base name and to that base name you can add a suffix and/or prefix(es). And in biology, you always have the first bit of the me which is the generic name and the second bit of the name which is the specific name. I know this probably wasn't very helpful but this is a very odd question if it's one that would be in an exam because chemistry and biology wouldn't usually be mixed together...

belka [17]3 years ago

6 0

Answer:

The similarity between the IUPAC naming and the zoological taxonomic nomenclature is that both are used for naming and providing identity.

The difference between the IUPAC naming and the zoological taxonomic nomenclature system is that the IUPAC naming is the process of naming the chemical compounds. These names are published in the Nomenclature of Organic Chemistry (Blue book). On the other hand, the zoological taxonomic naming system is used for naming the animals. The International Code of Zoological Naming has published many research works related to zoological naming.

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If 345.1 g of CO2 are placed in a vessel whose volume is 32.1 L at a temperature of 20.0oC, what will the pressure be? (R= 0.082

Finger [1]

Answer:

5.88atm

Explanation:

First, we obtain the number of mole of CO2 present in the vessel. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the question = 345.1g

Number of mole of CO2 =?

Number of mole = Mass/Molar Mass

Number of mole of CO2 = 345.1/44

= 7.84moles

Now we can easily calculate the pressure by doing the following:

Data obtained from the question include:

V (volume) = 32.1 L

T (temperature) = 20°C = 20 + 273 = 293K

R (gas constant) = 0.0821atm*L/mole*K

n (number of mole) = 7.84moles

P (pressure) =?

We will be making use of the ideal gas equation PV = nRT to calculate the pressure

PV = nRT

P = nRT/V

P = 7.84 x 0.0821 x 293/32.1

P = 5.88atm

Therefore, the pressure is 5.88atm

7 0

2 years ago

How many grams of NH3 are produced when 39.8 liters of N2 reacts?

algol13

Answer:

inj

Explanation:

3 0

2 years ago

Is MgCO3 an base or acid

lilavasa [31]

Ammonia is a (Lewis) base. Ammonium hydroxide (not ammonia) is a base. Ammonium hydroxide: NH4OH (ammonia gas dissolved in water) Ammonia: NH3

8 0

3 years ago

A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting i

Kisachek [45]

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

$\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}$

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

$\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}\\$

$\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J$\cdot$K$^{-1}$g$^{-1}$}\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J$\cdot$ K$^{-1}$g$^{-1}$}\times \Delta T_{\text{w}}\\\text{47.7 J$\cdot$K$^{-1}$}\times \Delta T_{\text{Cu}}& = &-\text{585 J$\cdot$ K$^{-1}$g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}$

The final temperature of the water would be 35.5 °C.

7 0

2 years ago

the average single and double n-o bond lengths are 136 pm and 122 pm, respectively. assuming these two structures were to make e

uranmaximum [27]

Resonance structures show that chemical species have equal bond lengths and bond angles when in resonance. The bond lengths of the N - O bond in HNO2 is 126pm.

Linus Pauling introduced the idea of resonance to explain the nature of bonding in compounds and ions where a single Lewis structure can not satisfactorily account for the observed properties of the chemical species.

If the two structures are resonance structures, they will have equal N - O bond lengths and bond angles. The bond length will be intermediate between that of a pure N - O single and N - O double bond. The actual N- O bond lengths in HNO2 are obtained as; 136 pm - 122 pm/2 = 126 pm.

Learn more: brainly.com/question/8155254

6 0

2 years ago