Ask question

Ask question

All categories

2 years ago

9

How does molecular shape affect polarity?

2 answers:

Andrej [43]2 years ago

6 0

Answer:

Letter D

Explanation:

stepladder [879]2 years ago

6 0

Answer:

An attraction between the positive end of one molecule and the negative end of another

Explanation:

You might be interested in

Which of the following would be a reasonable synthesis of CH3CH2CH2CH2OH?Group of answer choices

vampirchik [111]

Answer:

B. 1-Butene rightarrow (1) BH3: THF (2)H202, OH-

Explanation:

In the hydroboration of alkenes, an alkene is hydrated to form an alcohol with anti-Markovnikov orientation.

the reagent BH₃:THF is the way that borane is used in organic reactions. The BH₃ adds to the double bond of an alkene to form an alkyl borane. Peroxide hydrogen in basic medium oxidizes the alkyl borane to form an alcohol. Indeed, hydroboration-oxidation converts alkenes to alcohols by adding water through the double bond, with anti-Markovnikov orientation.

8 0

3 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

Using tongs place the 250 mL beaker on hot plate

LenKa [72]

Answer:

just use the tongs and put it on a plate

Explanation:

4 0

2 years ago

5.2 X 1026 molecules of CH4

olya-2409 [2.1K]

<h3 /><h3><u>Answer</u><u> </u><u>above</u><u> </u></h3>

<u>I</u><u> </u><u>cant</u><u> </u><u>send</u><u> </u><u>this</u><u> </u><u>with</u><u> </u><u>no</u><u> </u><u>text</u><u> </u>

<u>Have</u><u> </u><u>a</u><u> </u><u>nice</u><u> </u><u>day</u>

8 0

3 years ago

a compound with the empirical formula CH2 has a molar mass of 112 g/mol. What is the molecular formula for this compound?

miv72 [106K]

Hope it cleared your doubt :-D

5 0

3 years ago