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2 years ago

Which molecule is an alkene?

Chemistry

1 answer:

belka [17]2 years ago

4 0

D is an alkene because the ending is cis-2-pentene if it was an alkane is would be pentane.

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The state of matter with a specific volume but no definite shape is the ______. state

eimsori [14]

Answer:

C: liquid

Explanation:

Liquid has the set volume as it cannot be compressed or 'expanded' like a gas. However, liquids will take the shape of the container you put them in so their shape can change.

4 0

2 years ago

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What will be the bond angles for an sp3d2 hybridized atom participating in six identical single covalent bonds

Romashka [77]

Answer: electrons

Explanation:

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2 years ago

Calculate the percentrage of neutral form of aspirin that is present in the stomach ph 1.

Korolek [52]

Answer:

The 99.68% of the aspirin is present in the neutral form

Explanation:

Aspirin, Acetylsalicylic acid, is a weak acid with pKa = 3.5

Using Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the ionized form and HA the neutral form of the acid

Replacing with a pH of stomach of 1.0:

1.0 = 3.5 + log [A⁻] / [HA]

-2.5 = log [A⁻] / [HA]

3.16x10⁻³ = [A⁻] / [HA] (1)

A 100% of aspirin is = [A⁻] + [HA]

100 = [A⁻] + [HA] (2)

Replacing (2) in (1)

3.16x10⁻³ = 100 - [HA] / [HA]

3.16x10⁻³[HA] = 100 - [HA]

1.00316 [HA] = 100

[HA] = 99.68%

<h3>The 99.68% of the aspirin is present in the neutral form</h3>

8 0

3 years ago

What is the mass of Darmstadtium?

hammer [34]

Answer:

281 u

Explanation:

7 0

3 years ago

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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

8 0

3 years ago