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3 years ago

Select the true statements about hydrocarbons.

Chemistry

1 answer:

daser333 [38]3 years ago

8 0

Answer:

a. Hydrocarbons have low boiling points compared to compounds of similar molar mass.

b. Hydrocarbons are hydrophobic.

d. Hydrocarbons are insoluble in water.

Explanation:

As we know that the hydrocarbons is a mix of carbon and hydrogen. In this the availability of the electronegative atom is not there that shows there is no bonding of the hydrogen plus it is dissolved. Also, the hydrocarbons is considered to be a non-polar but as compared to the water, water is a polar

In addition to this, the strong bond is no existed that shows the lower boiling points

Therefore option A, B and D are right

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The partial pressure of cartbon dioxide in the atmgsphere is 0239 toer Caiculate the partial pressure in mm He and atm Rpund eac

Verizon [17]

Answer :

The pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.

The pressure of carbon dioxide in the atmosphere in atm is $3.14\times 10^{-4}atm$ .

Explanation :

The conversion used for pressure from torr to mmHg is:

1 torr = 1 mmHg

The conversion used pressure from torr to atm is:

1 atm = 760 torr

or,

$1torr=\frac{1}{760}atm$

As we are given the pressure of carbon dioxide in the atmosphere 0.239 torr. Now we have to determine the pressure of carbon dioxide in the atmosphere in mmHg and atm.

<u>Pressure in mmHg :</u>

As, $1torr=1mmHg$

So, $0.239torr=\frac{0.239torr}{1torr}\times 1mmHg=0.239mmHg$

Thus, the pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.

<u>Pressure in atm:</u>

As, $1torr=\frac{1}{760}atm$

So, $0.239torr=\frac{0.239torr}{1torr}\times \frac{1}{760}atm=3.14\times 10^{-4}atm$

Thus, the pressure of carbon dioxide in the atmosphere in atm is $3.14\times 10^{-4}atm$ .

3 0

2 years ago

How many naturally occurring elements are present on the product table?

algol [13]

The first 92 elements of the periodic table are naturally occurring elements.

6 0

3 years ago

A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at

weqwewe [10]

Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of tin = $m_1=18.5 g$

Specific heat capacity of tin = $c_1=0.21 J/g^oC$

Initial temperature of the tin = $T_1=97.38^oC$

Final temperature = $T_2$ =T

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.7 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=21.52^oC$

Final temperature of water = $T_2$ =T

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)$

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

5 0

3 years ago

How many significant figures are needed in the answer when 1.31m is multiplied by 6.5 m

dmitriy555 [2]

Answer:

two

Explanation:

The number of significant figures needed in the answer is 2.

This is because when finding the products of two numbers, the result is as accurate as the least number of significant figures of the numbers being multiplied.

Here the numbers being multiplied are;

1.31m

6.5m

1.31m has 3 significant figures

6.5m has 2 significant figures.

So, the product will have 2 significant figures

6 0

2 years ago

Calculate δs∘rxn for the reaction2no(g) o2(g)→2no2(g)express your answer to one decimal place and include the appropriate units

Anna35 [415]

The δs∘rxn for the reaction $2NO(g) + O_{2}$ → $2NO_{2} (g)$ will be -146 J/K.

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Entropy would be a measurement of the system's unpredictability or disorder. The entropy increases as randomness do. It has broad properties as well as a state function. It has the unit $JK^{-1} mol^{-1}$ .