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3 years ago

!! ANSWER QUICK !!

Chemistry

2 answers:

sammy [17]3 years ago

8 0

Answer:

They are correct!!!

Explanation:

kari74 [83]3 years ago

3 0

Answer:

I can use a dichotomous key. It helps me classify objects by sorting it out with "yes" and "no" questions.

I can use a Punnett Square. It helps me classify what genes the offspring will receive simply by figuring out the recessive and dominant genes as well as the hetzygous and homzygous.

Now give an example of which ever chart you choose by drawing it if that is required. For the Punnett Square label each of the squares Top right Hetzygous, top left dominant, bottom left recessive, bot-tom right homzygous. And for the dichotomous key put a 5-7 length branch showing the animals that have fur, can breathe under water, what cannot or doesn't have those traits. or something similar

Hopefully this helps :)

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A gallon of paint covers 350 square feet.

trapecia [35]

Answer:

2,137... gallons

Explanation:

(160+140+14x16x2)/350=2,137...

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Which of the following would not be defined as an acid A.KOH, B. HNO3, D. H3PO4 and why

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3 years ago

Electrochemistry - Equilibrium

Ipatiy [6.2K]

Answer:

Explanation:

The relation between equilibrium constant and Ecell is given below .

E⁰cell = (RT / nF ) lnK , F is faraday constant T is 273 + 25 = 298 K

E⁰cell = 1.46 - 1.21 = .25 V

n = 2

Putting the values

.25 = (8.314 x 298 lnK) / (2 x 96485 )

lnK = 19.47

K = 2.85 x 10⁸

2 )

Change in free energy Δ G

Δ G ⁰ = nE⁰ F

n = 4

E⁰ = .4 + .83 = 1.23 V

Δ G ⁰= 4 x 1.23 x 96485

= 474706 J / mol

3 )

E⁰cell = (RT / nF ) lnK

n = 2

1.78 = 8.314 x 298 lnK / 2 x 96485

lnK = 138.638

K = 1.62 x 10⁶⁰

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3 years ago

In a titration experiment a student uses 1.4 m hbr solution and the indicator phenolphthalein to determine the concentration of

Likurg_2 [28]

The question is incomplete. Complete question is attached below:
...........................................................................................................................

Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL

We know that, M1V1 = M2V2
(HBr) (KOH)

Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
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Concentration of KOH is 0.9756 M.

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