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Karo-lina-s [1.5K]
3 years ago
12

Please I really need help:

Chemistry
2 answers:
12345 [234]3 years ago
6 0
The answer is C I think
puteri [66]3 years ago
4 0
<span>C. 159.70 Fe=55.85g/mol * 2 = 111.70 O=16.00g/mol * 3 = 48.00 +_______ 159.70</span>
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Napisz równania reakcji chemicznych w formie cząsteczkowej, jonowej i jonowej skróconej. Podkreśl wzory soli trudno rozpuszczaln
Ierofanga [76]

Answer:idioma equivocado ????

Explanation:

3 0
3 years ago
Hydrogen bonds are formed between molecules when hydrogen is covalently bonded to an element that has a
stepan [7]

Answer:

choice 3

Explanation:

Hydrogen bonding occurs when a H is bonded to a highly electronegative atom like F. Towards the upper right corner, most atoms have high electronegativity and small atomic radii, according to periodic trends.

Please give thanks :)

7 0
3 years ago
As the temperature of a fixed volume of a gas increases the pressure wil
Sati [7]

Answer:

As the temperature of a fixed volume of a gas increase the pressure will increase.

Explanation:

According to the Gay- Lussac's Law,

" The pressure of given amount of gas is directly proportional to the temperature at a constant volume"

Mathematical expression:

           P ∝ T

          P = CT

          P / T = C

As the temperature increase, the pressure also increase.

The initial and final expression of volume and pressure can be written as,

P₁ / T₁  = P₂ / T₂

3 0
3 years ago
A 15 kg television sits on a shelf at a height of 0.3. How much gravitational potential energy is added to the television when i
slavikrds [6]

Answer:

Explanation:

Answer:

Explanation:

Answer:

Explanation:

Answer:

Explanation:

Answer:

Explanation:

7 0
2 years ago
The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse
Umnica [9.8K]
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
5 0
3 years ago
Read 2 more answers
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