3.How are climate and weather different?

A mains radio turns ________ energy into ______ energy and ______ energy.

liberstina [14]

Shooting a rubber band turns _________ energy into _______ energy Elastic potential energy into mechanical energy

The ___________ energy from the sun transforms into ____________ energy in plants

Radiant energy into chemical energy

A runner turns _______ energy into ________ energy

Chemical energy into mechanical energy

A hand crank radio turns _____ energy into ______ energy into _____________ energy

Mechanical energy into electrical energy intosound energy

A nuclear power plant turns ______ into ______

Nuclear energy into electrical energy

A fan turns ______ energy into ______ energy

electrical energy into mechanical energy

A flashlight turns _____ energy into ________ energy into ________ energy and ________ energy

Chemical energy into electrical energy into light and thermal energy

A candle turns _______ energy into ___________ energy and ____________ energy

Chemical energy into light and thermal energy

A television uses ___________ energy and transforms it into ___________, ______________, and ____________

Electrical energy into thermal, light and sound energy

Falling water in a dam has the following energy transformations: ______________ to _____________ to _______________ to _____________

Kinetic energy in falling water from mechanical to electrical to light

3 0

2 years ago

A buffer is prepared such that [H2PO4-] = 0.095M and [HPO42-] = 0.125M? What is the pH of this buffer solution? (pKa = 7.21 for

ANTONII [103]

Answer:

pH of the buffer is 7.33

Explanation:

The mixture of the ions H₂PO₄⁻ and HPO₄²⁻ produce a buffer (The mixture of a weak acid, H₂PO₄⁻, with its conjugate base, HPO₄²⁻).

To find pH of a buffer we use H-H equation:

pH = pka + log [A⁻] / [HA]

<em>Where A⁻ is conjugate base and HA weak acid.</em>

<em />

For the H₂PO₄⁻ and HPO₄²⁻ buffer:

pH = pka + log [HPO₄²⁻] / [H₂PO₄⁻]

Computing values of the problem:

pH =7.21 + log [0.125M] / [0.095M]

pH = 7.33

<h3>pH of the buffer is 7.33</h3>

<em />

4 0

3 years ago

Air pressure is greatest near the surface of the earth near the _

rewona [7]

Answer:

With greater depth of the atmosphere, more air is pressing down from above. Therefore, air pressure is greatest at sea level and falls with increasing altitude

Explanation:

On top of Mount Everest, which is the tallest mountain on Earth, air pressure is only about one-third of the pressure at sea level.

6 0

3 years ago

Read 2 more answers

A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to

Snezhnost [94]

Explanation:

It is known that $pK_{a}$ of propionic acid = 4.87

And, initial concentration of propionic acid = $\frac{0.19}{1.20}$

= 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

= 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence, pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.216}{0.158}$

= 4.87 + log (1.36)

= 5.00

Therefore, when 0.02 mol NaOH is added then,

Moles of propionic acid = 0.19 - 0.02

= 0.17 mol

Hence, concentration of propionic acid = $\frac{0.17}{1.20 L}$

= 0.14 M

and, moles of sodium propionic acid = (0.26 + 0.02) mol

= 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.28 mol}{1.20 L}$

= 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.23}{0.14}$

= 4.87 + log (1.64)

= 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

Therefore, when 0.02 mol HI is added then,

Moles of propionic acid = 0.19 + 0.02

= 0.21 mol

Hence, concentration of propionic acid = $\frac{0.21}{1.20 L}$

= 0.175 M

and, moles of sodium propionic acid = (0.26 - 0.02) mol

= 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.24 mol}{1.20 L}$

= 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.2}{0.175}$

= 4.87 + log (0.114)

= 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0

3 years ago

5.36 liters of nitrogen gas are at STP. What would be the new volume if we increased the moles from 3.5 moles to 6.0 moles?

Aleksandr [31]

Answer:

$V_2=9.20L$

Explanation:

Hello there!

In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

$\frac{V_2}{n_2}= \frac{V_1}{n_1}$

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

$V_2= \frac{V_1n_2}{n_1} \\\\V_2=\frac{5.36L*6.0mol}{3.5mol}\\\\V_2=9.20L$

Best regards!

3 0

3 years ago