Element Name: Oxygen,

Be sure to answer these questions in your journal entry: • How long was the journey to Earth? • How fast did you travel? • Was y

damaskus [11]

Answer: What you doing

Explanation:

3 0

2 years ago

When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant so

zmey [24]

Answer: 54.4 kJ/mol

Explanation:

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 mole

Now we have to calculate the mass of water:

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $50ml+50ml=100ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 100 g

$T_{final}$ = final temperature of water = $27.5^0C$

$T_{initial}$ = initial temperature of metal = $21.0^0C$

Now put all the given values in the above formula, we get:

$q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C$

$q=2719.6J=2.72kJ$

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of $HCl$ :

0.05 moles of $HCl$ releases heat = 2.72 KJ

1 mole of $HCl$ releases heat = $\frac{2.72}{0.05}\times 1=54.4KJ$

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

6 0

2 years ago

Read 2 more answers

What is the name of this hydrocarbon? 2-dimethylyne 2-methylbutane 4-methylbutene 4,3-methylbutyne 4,4-methylbutane

Alisiya [41]

You have shown no structure, still we can work out for the answer :).

First Option: 2-dimethylyne: This name is incorrect, because it does not contain any parent chain name.

Second Option: 2-methylbutane: I have drawn its structure below, compare it with your structure.

Third Option: 4-methylbutene: I have drawn the structure for this name. But this name is also against IUPAC rules, and its correct name is 1-Pentene.

Fourth Option: 4,3-methylbutyne: Again incorrect name, 4,3 are two positions, but here only one substituent is given (methyl). I have drawn structure of 3,4-Dimethylbutene, this name is incorrect, and the correct name for this compound is 3-Methyl-1-pentene.

Fifth Option: 4,4-methylbutane: Again incorrect name, 4,4 means at 4 position two substituents, but in name only methyl is given. Anyhow lets make it two and draw a structure, for 4,4-Dimethylbutane. Ooops!! This name is also incorrect, and the correct name for this compound is 2-Methylpentane.

Result:
Among all options only, option B (2-Methylbutane) is correctly named, and this will be the correct name for the structure given to you.

8 0

3 years ago

Read 2 more answers

Draw the most stable resonance structure for the intermediate in the electrophilic aromatic bromination of aniline, anisole, and

ASHA 777 [7]

Answer:

Here's what I get

Explanation:

(a) Intermediates

The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).

(b) Relative Stabilities

The relative stabilities decrease in the order shown.

N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.

(c) Relative reactivities

The relative reactivities would be

C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃

4 0

3 years ago

A gas is at a pressure of 3.70 atm. What is this pressure in kilopascals?

Yanka [14]

Answer:
= 374.90 kPa

Calculation:
As we know atm and kiloPascal are related to each other as,

1 atm = 101.325 kPa
So,
3.70 atm = X
Solving for X,
X = (3.70 atm × 101.325 kPa) ÷ 1 atm

X = 374.90 kPa

7 0

3 years ago