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3 years ago

7

A volcano violently erupts near a small town and wildlife preserve. Which effect might occur? Some local species develop new ada

ptations. New species form at the wildlife preserve. Many wildlife species become extinct. Ash and dust fill the air and reduce sunlight.

2 answers:

Anna35 [415]3 years ago

4 0

Answer:

D

Explanation:

NemiM [27]3 years ago

3 0

Answer:

Ash and dust fill the air and reduce sunlight.

Explanation:

Ash and dust comes out of volcanoes, and it is bad for the environment. Polluting the air and causing fogginess, it also reduces sunlight.

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A 25.00 mL sample of vinegar was titrated with 39.27 mL of 0.4293 M NaOH. Calculate the concentration of acetic acid in the vine

QveST [7]

Answer:

0.6743 M

Explanation:

HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O

First we <u>calculate how many NaOH moles reacted</u>, using the <em>definition of molarity</em>:

Molarity = moles / volume

moles = Molarity * volume

0.4293 M * 39.27 mL = 16.86 mmol NaOH

<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.

Finally we <u>calculate the concentration (molarity) of acetic acid</u>:

16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M

4 0

3 years ago

"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum

UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = $25^{o} C$

= (25 + 273) K

= 298.15 K

Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

0.698 = $C \times 0.082 \times 298.15 K
$

C = 0.0285

This also means that,

$\frac{\text{moles}}{\text{volume (in L)}}$ = 0.0285

So, moles = 0.0285 × volume (in L)

= 0.0285 × 0.100

= $2.85 \times 10^{-3
}$

Now, let us assume that mass of $C_{12}H_{23}O_{5}N$ = x grams

And, mass of $C_{12}H{22}O_{11}$ = (1.00 - x)

So, moles of $C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}$

= $\frac{x}{369}$

Now, moles of $C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}$

= $\frac{x}{369} + \frac{(1.00 - x)}{342}$

= $2.85 \times 10^{-3}$

= x = 0.346

Therefore, we can conclude that amount of $C_{12}H_{23}O_{5}N$ present is 0.346 g and amount of $C_{12}H_{22}O_{11}$ present is (1 - 0.346) g = 0.654 g.

4 0

3 years ago

Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+

eimsori [14]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

$X\rightarrow X^{n+}+ne^-$

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

<u>Option a:</u> $Ag^+$

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

<u>Option b:</u> $Pb$

This metal can easily get oxidized to $Pb^{2+}$ ion and the standard oxidation potential for this is 0.13 V

$Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V$

<u>Option c:</u> $H_2$

This metal can easily get oxidized to $H^{+}$ ion and the standard oxidation potential for this is 0.0 V

$H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V$

<u>Option d:</u> $Ag$

This metal can easily get oxidized to $Ag^{+}$ ion and the standard oxidation potential for this is -0.80 V

$Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V$

<u>Option e:</u> $Mg^{2+}$

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive $E^o$ potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

8 0

3 years ago

Analyze your results.

raketka [301]

#b

According to Le C ha.te llors principle of we increase concentration of reactants or products equilibrium shifts.

Yes increases

#c

Rate of reaction also increases

#e

Stated in b

3 0

2 years ago

When naming a binary compound, what ending do you use to represent anions?

spin [16.1K]

I believe You replace the ending of the elements name with -ide. example: magnesium flourine should should be magnesium flouride.

6 0

3 years ago