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Phantasy [73]
3 years ago
11

The ionization energy of an element is

Chemistry
1 answer:
aalyn [17]3 years ago
6 0
The ionization energy or ionization potential is - the energy required to remove an electron from an element in its gaseous state.
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USPshnik [31]
<span> The temperature of the water in both beakers is greater than 8 °C
This is because since the water passes on heat to the spheres, that means they are at a higher temperature then the spheres. if they were at a lower temp. The spheres would pass on heat to the water.
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3 0
3 years ago
WILL GIVE FREE BRAINLIEST EVERYDAY! DO NOT look up answers!! PLEASE FOLLOW DIRECTIONS ON FUTURE QUESTIONS OR WRONG!!! (Branliest
Maslowich

Answer:

  • <u>Purpose of Iodine:</u>

Iodine plays a vital role in thyroid health. Our thyroid gland, which is located at the base of the front of your neck, helps regulate hormone production. These hormones control your metabolism and heart health.

  • <u>Symbol:</u>

The symbol of Iodine is " I ".

  • <u>Atomic Mass:</u>

Atomic mass of Iodine is 126.90447 u

  • <u>Protons:</u>

No. of protons in Iodine is 53.

  • <u>Neutrons:</u>

No. of neutrons in Iodine is 74.

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8 0
3 years ago
If the molar heat of fusion of water is 6.01
Zolol [24]

Answer : The heat energy required to melt 2 kg of ice was, 667.7 kJ

Explanation :

First we have to calculate the moles of ice.

\text{Moles of ice}=\frac{\text{Given mass ice}}{\text{Molar mass ice}}=\frac{2kg}{18g/mol}=\frac{2000g}{18g/mol}=111.1mol

Now we have to calculate the heat energy.

As, heat energy required to melt 1 mole of ice = 6.01 kJ

So, heat energy required to melt 111.1 mole of ice = 111.1 × 6.01 kJ

                                                                                  = 667.7 kJ

Therefore, the heat energy required to melt 2 kg of ice was, 667.7 kJ

6 0
4 years ago
As a part of a clinical study, a pharmacist is asked to prepare a modification of a standard 22g package of a 2% mupirocin ointm
Vesna [10]

Answer:

226.8 mg of mupirocin powder are required

Explanation:

Given that;

weight of standard pack = 22 g

mupirocin by weight = 2%

so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g

so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment

mg of mupirocin powder are required = ?, lets rep this with x

Total weight of ointment = 22 + x g

Amount of mupirocin  = 0.44 + x g

percentage of mupirocin  in ointment is 3?

so

3/100 = 0.44 + x g / 22 + x g

3( 22 + x g ) = 100( 0.44 + x g )

66 + 3x g = 44 + 100x g

66 - 44 = 100x g - 3x g

97 x g = 22

x g = 22 / 97

x g = 0.2268 g

we know that; 1 gram = 1000 Milligram

so 0.2268 g = x mg

x mg = 0.2268 × 1000

x mg  = 226.8 mg

Therefore, 226.8 mg of mupirocin powder are required

8 0
3 years ago
Which Domains would you find single cell organisms
kogti [31]
Archaea and Bacteria
8 0
3 years ago
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