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3 years ago

15

this environment has lots of trees high tenpatures and high mumidity all year long which environment is this describing Grasslan

d, Rainforest, Swamp, Wetland

1 answer:

Genrish500 [490]3 years ago

7 0

I believe the answer is Rainforest.

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Metals have low ionization energies and readily share their ________ or outer electrons with each other to form an electron ____

Dmitry [639]

Explanation:

Metals are the species which readily lose electrons in order to attain stability. This electron lost by the atom is actually present in its outermost shell which is also known as valence shell.

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period.

But when we move down a group then there occurs an increase in atomic size of the atoms due to addition of number of electrons in the atoms. Hence, ionization energy decreases along a group.

Thus, we can conclude that metals have low ionization energies and readily share their valence or outer electrons with each other to form an electron sea. These electrons are delocalized or shared among all the atoms that are bonded together and can therefore move freely throughout the metal structure.

3 0

3 years ago

Read 2 more answers

A block of ice is cooled from -0.5 degrees Celsius to -10.1 degrees Celsius. Calculate the temperature change, delta T, in degre

ss7ja [257]

Delta T= T final - T initial
Tfinal= -101.1 °C
Tinitial= -0.5 °C
•Delta T = -101.1°C - (-0.5°C)
=100.6°C

Kelvin= °C + 273
= -100.6 + 273
= 172.4 Kelvin

3 0

3 years ago

If the density of gold is 19.3 g/cm3 what would be the volume of 550 g of gold?

RUDIKE [14]

Answer:

28.497 cm3

Explanation:

Formula

D=m/v

Given data:

density = 19.3g/cm3

mass = 550 g

Now we will put the values in formula:

V=m/d

V=550 g/ 19.3 g/cm3 = 28.497 cm3

So the volume of gold is 28.497 cm3.

7 0

3 years ago

A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Also, relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0

3 years ago

1. Calculate the mass (g), of 4.5 atoms of oxygen.<br> I

konstantin123 [22]

Answer

thus, 4 moles of oxygen gas (O2) would have a mass of 128 g.

3 0

3 years ago