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Alenkasestr [34]

3 years ago

15

Large-scale environmental catastrophes _______.

1 answer:

Ymorist [56]3 years ago

3 0

Answer:

D. All of the Above

Explanation:

i just took the test on edgenuity

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____________ is a point source of water pollution; _____________ is a nonpoint source of water pollution.

Deffense [45]

Answer:

good afternoon

Explanation:

thanks I will send

6 0

2 years ago

How many liters of a 1.5 M solution can you make if you have .50 mol of KCl?

vaieri [72.5K]

Answer:

The answer to your question is 0.33 liters

Explanation:

Data

Volume = ?

Molarity = 1.5 M

number of moles = 0.5

Formula

$Molarity = \frac{number of moles }{volume}$

Solve for V

$Volume = \frac{number of moles}{molarity}$

Substitution

$Volume = \frac{0.5}{1.5}$

Simplification and result

Volume = 0.33 l

7 0

3 years ago

According to newtons third law forces always occur in equal but____ pairs?

arsen [322]

'Every action has an equal but opposite reaction' They are equal, but opposite.

8 0

3 years ago

3. Burns from boiling water can be severe, caused by the transfer of energy from the boiling water to the

Free_Kalibri [48]

Answer:

Q = 3937.56 J

Explanation:

Heat transferred due to change in temperature is given by :

$Q=mc\Delta T$

c is the specific heat of water, c=4.18 J/g-°C

We have, m = 15 g, $T_i=100^{\circ} C\ \text{and}\ T_f=37.2^{\circ} C$

So,

$Q=15\times 4.18\times (37.2-100)\\Q=-3937.56\ J$

Hence, 3937.56 J of heat is transferred.

8 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 :

<u>(a) a solution of pH 3.0</u>

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, $1.0\times 10^{-11}M$

<u>(b) a solution of 0.10 M $NH_3$ </u>

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

<u>(c) a solution with a pOH of 12.</u>

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, $1.0\times 10^{-12}M$

<u>(d) a solution of 0.10 M NaOH</u>

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

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

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