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3 years ago

If snails and crayfish die, what happens to the other species?

Chemistry

2 answers:

Pani-rosa [81]3 years ago

8 0

Answer:

Other species will slowly die, because the species that eat snails and crayfish will not have anything to eat, then the thing that eats them will not have anything. This process will go on and on. So long story short populations will decrease, then off.

elixir [45]3 years ago

4 0

Crayfish are very intolerant of pollution and other human-generated fouling of their environment. ... This abundance may be due more to the acid-generated absence of fish which prey on crayfish than to a direct, positive influence of the acid on the crayfish itself.

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What is the pH of a solution with a H+ ion of 2×10-¹²?

kap26 [50]

Answer:

11.7

Explanation:

The pH is the negative logarithm of the concentration of H+ ions. If the concentration is 2×10-¹² the the pH will be -log(2×10-¹²) which is 11.698 which can be round up to 11.7.

5 0

3 years ago

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

(b) a solution of 0.10 M $NH_3$

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

(c) a solution with a pOH of 12.

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

(d) a solution of 0.10 M NaOH

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

(e) a $1\times 10^{-4}M$ solution of $HNO_2$

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

3 years ago

A gas sample occupies 3.25 liters at 297.5K and 2.4 atm. Determine the temperature at which the gas will occupy 4.25 L at 1.50 a

lorasvet [3.4K]

For equal moles of gas, temperature can be calculated from ideal gas equation as follows:

P×V=n×R×T ...... (1)

Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.

2.4 atm ×3.25 L=n×R×297.5 K

Rearranging,

n\times R=0.0262 atm L/K

Similarly at final pressure and volume from equation (1),

1.5 atm ×4.25 L=n×R×T

Putting the value of n×R in above equation,

1.5 atm ×4.25 L=0.0262 (atm L/K)×T

Thus, T=243.32 K

7 0

3 years ago

A natural atom possesses the atomic number of 13 and a atomic mass of 27. Three electrons are lost. From what region of the atom

Anarel [89]

Answer:

The electrons are lost from the valence shell (outermost electron shell) of the atom.

Explanation:

This is able to be inferred not only because valence electrons being lost first is a trend but also because the atom in question has actually 3 valence electrons (13-2-8 = 3).

7 0

3 years ago

2) what is the mass of 6.02 x 1023 atoms of arsenic?

zzz [600]

Data:
Arsenic Molar Mass = 74,9216 ≈ 75 u (atomic mass unit)

Solving:

1 mole of arsenic → 75g ------------ 6,02*10²³ molecules
..................................X -------------- 1 molecule

6,02*10²³X = 75
$x = \frac{75}{6,02*10^{23}}$
$\boxed{x \approx 1,24*10^{24}grams}$

6 0

3 years ago