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Artyom0805 [142]
3 years ago
10

During the formation of a chemical bond between two hydrogen atoms, which of the following statements is always true?

Chemistry
1 answer:
Wewaii [24]3 years ago
5 0

A) Energy is released during the formation of the bond.

Explanation:

During the formation of a chemical bonds between two hydrogen atoms, energy is always released during the formation of this bond type.

Bond formation process is usually exothermic and energy is released during the formation of the bond.

  • Bond breaking process is an endothermic process in which energy is absorbed from the surrounding.
  • Whenever a bond is broken, the bond energy value is positive but when a bond is formed, the bond energy value is given a negative sign.

For a bond formation process in which hydrogen atoms are bonded covalently, energy is usually released.

Learn more:

Enthalpy changes brainly.com/question/10567109

#learnwithBrainly

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Sea water contains roughly 158.0g of Nacl per liter. What is the molarity of sodium chloride in sea water?
IrinaK [193]
Molar mass NaCl = 58.5 g/mol

C = 158.0 g/L

Molarity =  C / molar mass

M = 158.0 / 58.5

M = 2.7000 M

hope this helps!
6 0
3 years ago
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2
Arlecino [84]

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide  =\dfrac{70.4 \ g}{121.14 \ g/mol}

= 0.58 mole

The molality = \dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent  }

= \dfrac{0.58 }{0.85 }

= 0.6837

Using the formula:

\mathbf {dT  = l   \times  k_f  \times m}

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf ×  0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl = \dfrac{70.4 \ g}{53.491 \  g /mol}

= 1.316 mol

The molality = \dfrac{1.316 \ mol}{0.85 \ kg}

= 1.5484

Thus;

the above kf value is used in determining the  Van't Hoff factor for  NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

l = \dfrac{9.9}{3.949 \times 1.5484 \ m}

l = 1.62

5 0
3 years ago
How many atoms are contained in 5.77 grams of aluminum? Express your answer in exponential notation (Ex: 6.02*1023 would be ente
bagirrra123 [75]

Answer:

5.09

Explanation:

6 0
3 years ago
A vial containing radioactive selenium-75 has an activity of 3.0 mCi/mL. If 2.6 mCi are required for a leukemia test, how many m
oksian1 [2.3K]

Answer : The 866.66\mu L must be administered.

Solution :

As we are given that a vial containing radioactive selenium-75 has an activity of 3.0mCi/mL.

As, 3.0 mCi radioactive selenium-75 present in 1 ml

So, 2.6 mCi radioactive selenium-75 present in \frac{2.6mCi}{3.0mCi}\times 1ml=0.86666ml\times 1000=866.66\mu L

Conversion :

(1ml=1000\mu L)

Therefore, the 866.66\mu L must be administered.

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