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mrs_skeptik [129]
3 years ago
7

What is one major purpose for an animal to produce unusual sounds, display bright

Chemistry
1 answer:
bezimeni [28]3 years ago
7 0
I would say courtship, we this is when an animal would be looking for a mate. They would want to make themselves different and bright colors could attract a possible mate for them. I hope this helps!
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Magnesium has three naturally occurring isotopes(Mg-24, Mg-25, and Mg-26). The atomic mass and natural abundance of Mg-24 are 23
xxMikexx [17]

Answer:

The answer to your question is below

Explanation:

Data

               Atomic mass               Abundance

Mg -24 =    23.9850                         79%

Mg -25 =    24.9858                         10%  

Mg - 26=          ?                                  ?

Process

1.- Find the abundance of Mg-26

The sum of the abundance of the three isotopes equals 100%

         

 Abundance Mg-24 + Abundance-Mg-25 + Abundance-26 = 100

            79 + 10 + Abundance-26 = 100

            Abundance Mg-26 = 100 -89

           Abundance Mg-26 = 11%

2.- Find the atomic mass of Mg-26, the average atomic mass is 24.31 amu

Average atomic mass = (abundance)(atomic mass- 24) + (abundance)

                                         (atomic mass-25) + (abundance)(atomic mass-26)

    24.31 = (0.79)(23.9850)+ (0.10)(24.9858) + (Atomic mass-26)(0.11)

    24.31 = 18.94815 + 2.4986 + 0.11(atomic mass-26)

    24.31 = 21.4342 + 0.11(atomic mass- 26)

    0.11(atomic mass-26) = 24.31 - 21.4342

    0.11(atomic mass-26) = 2.8758

    Atomic mass-26 = 2.8758/0.11

 <u>   Atomic mass-26 = 26.1436 amu</u>

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3 years ago
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Describe Write your own caption for this photo.
koban [17]

Answer:

Sit by the fire to warm up

Explanation:

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2 years ago
Heating 235 g of water from 22.6°C to 94.4°C in a microwave oven requires 7.06 × 104 J of energy. If the microwave frequency is
Darya [45]

Answer: 3.69 × 10^27

Explanation:

Amount of energy required = 7.06 × 10^4 J

Frequency of microwave (f) = 2.88 × 10^10 s−1

Planck's constant (h) = 6.63 × 10^-34 Jᐧs/quantum

Recall ;

Energy of photon = hf

Therefore, energy of photon :

(6.63 × 10^-34)j.s× (2.88 × 10^10)s^-1

= 19.0944 × 10^(-34 + 10) = 19.0944×10^-24 J

Hence, number of quanta required :

(7.06 × 10^4)J / (19.0944 × 10^-24)J

= 0.369 × 10^(4 + 24) = 0.369×10^28

= 3.69 × 10^27

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