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shtirl [24]
3 years ago
13

WILL GIVE BRAINLIEST NO FAKE ANSWERS PLEASE!

Chemistry
2 answers:
arlik [135]3 years ago
5 0

Answer:

3.2 grams

Explanation:

Took the test and got it correct

AleksAgata [21]3 years ago
3 0

Answer:

A) (3.2g)

Explanation:

Did you reposed this? Because I remember answering this

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Which would easily give up more electrons during bonding?
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Lie I think I believe
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Describe each highlighted bond in terms of the overlap of atomic orbitals. (If the highlighted bond is not a pi bond, select the
Anestetic [448]

The image of the bonds are missing, so i have attached it.

Answer:

A) - Sigma bond

-Sp³ and Sp³

- None

B) - Sigma and pi bond

- Sp² of C and p of O

- p of C and P of O

Explanation:

A) For compound 1;

- the molecular orbital type is sigma bond due to the end-to-end overlapping.

- Atomic orbitals in the sigma bond will be Sp³ and Sp³

- Atomic orbitals in the pi bond would be nil because there is no pi bond.

B) For compound 2;

- the molecular orbital type is sigma and pi bond

-Atomic orbitals in the sigma bond would be Sp² of C and p of O

- The Atomic orbitals in the pi bond will be; p of C and p of O

6 0
3 years ago
Suppose you are titrating an acid solution with a base solution of known concentration. To calculate the concentration of the ac
Arturiano [62]

Answer:

i would say use 2

Explanation:

4 0
3 years ago
I'm currently doing a Castle Learning part teacher and it's graded
kipiarov [429]

See the image below.

An excited electron is in a <em>high-energy state</em>.

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4 0
3 years ago
A total of 663 cal 663 cal of heat is added to 5.00 g 5.00 g of ice at − 20.0 °C . −20.0 °C. What is the final temperature of th
Serjik [45]

Answer:

Final Temperature = 305.9 K or 32.9°C

Explanation:

Information given;

Mass (m) = 5g

Amount of Heat (H) = 663 calories

Initial Temperature = -20°C + 273 = 253K (Converting to Kelvin Temperature)

Final Temperature?

Specific Heat capacity of Water (c) =?

One might probably rush to use H = m * c * (T2-T1) to calculate the fianl temperature. That would have been wrong because Ice would melt when heat is being applied to it. And using that formular directly would lead to not considering the amount of heat required to melt it.

First let us check the amount of heat required to raise the temperature of the ice to 0°C

In this case now; Final temperature = 0°C + 273 = 273 K (Converting to Kelvin)

H = m * c * (T2-T1)

H = 5 * 1 * (273 - 253)

H = 5 * 1 * 20 = 100 cal

This shows the heat supplied is enough (663 cal is more than 100 cal) to bring the ice to its melting point.

Let's see if it would be sufficient to melt it.

Amount of Heat required to Melt ice;

H = mL;

where L = heat of fusion = 79.7 cal/g

H = 5 * 79.7 = 398.5 cal

Again, the heat is sufficient to melt it; the remaining heat would be used in raising the temperature of the liquid water.

In this case, initial temperature = 0°C + 273 = 273 K (Converting to Kelvin)

Amount of Heat left = 663 - 398.5 - 100 = 164.5 cal

Final temperature is given as;

H = m * c * (T2-T1)

164.5 = 5 * 1 * (T2 - 273)

T2 - 273 = 164.5 / 5

T2 = 273 + 32.9 = 305.9 K

Final Temperature = 305.9 K or 32.9°C

4 0
3 years ago
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