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Artemon [7]
2 years ago
14

What power of ten is one light year?

Chemistry
1 answer:
Sav [38]2 years ago
8 0

Answer:

9.46x10^12 kilomiters is your answer

Explanation:

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For each of the following ionic bonding examples, show the transfer of electrons, the ions resulting with the charges and the fi
fgiga [73]

Answer:

BaO

  • Ba gives up two electrons and becomes Ba⁺²
  • O takes two electron and becomes O⁻²

Al₂O₃

Mg₃P₂

Na₂O

CaCl₂

6 0
3 years ago
An exothermic reaction is what<br> energy.
leonid [27]

Answer:

An exothermic reaction releases energy. Endothermic takes in energy.

Explanation:

7 0
3 years ago
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A 453 g piece of glass at 25.7∘C is left outside on a sunny day. How much heat must the glass absorb from the sun in order to re
Tju [1.3M]

Answer:

Q = 5555.6J

Explanation:

Mass of glass piece, m = 453g

initial temperature = 25.7°C

temperature to be attained = 40.3°C

⇒change in temperature, Δt = 40.3 - 25.7 = 14.6°C

specific heat of glass, s = 0.840J/g°C

Heat absorbed, Q = msΔt

⇒Q = 453×0.840×14.6 = 5555.592J

⇒<u>Q = 5555.6J</u> (rounded to the nearest tenth)

7 0
4 years ago
5. an atom that has lost or gained an electron and has become charged
densk [106]

Answer:

It would be compound.

Explanation:

It is this way because if it adds another proton it becomes more positive that nuetral, and if you add an electron it just makes the atom more dense. That is why the answer is compound. Hope this helped :)

5 0
3 years ago
A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression o
Ludmilka [50]

Answer:

0.297 °C

Step-by-step explanation:

The formula for the <em>freezing point depression </em>ΔT_f is

ΔT_f = iK_f·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For glucose,

       glucose(s) ⟶ glucose(aq)

1 mole glucose ⟶ 1 mol particles     i = 1

Data:

Mass of glucose = 10.20 g

  Mass of water = 355 g

                 ΔT_f = 1.86 °C·kg·mol⁻¹

Calculations:

(a) <em>Moles of glucose </em>

n = 10.20 g × (1 mol/180.16 g)

  = 0.056 62 mol

(b) <em>Kilograms of water </em>

m = 355 g × (1 kg/1000 g)

   = 0.355 kg

(c) <em>Molal concentration </em>

b = moles of solute/kilograms of solvent

  = 0.056 62 mol/0.355 kg

  = 0.1595 mol·kg⁻¹

(d) <em>Freezing point depression </em>

ΔT_f = 1 × 1.86 × 0.1595

        = 0.297 °C

3 0
3 years ago
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