1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
amid [387]
3 years ago
9

Given the balanced equation:

Chemistry
1 answer:
Marta_Voda [28]3 years ago
7 0
The answer is (2) A bond is formed and energy is released. The left side of equation is I atom and the right side of equation is I2 molecule. So the bond is formed between I atom to form I2 molecule. And forming bond will release energy while breaking bond will absorb energy.
You might be interested in
Model describe how you would show j.j. thomson's model of the atom using small beads and clay
faltersainse [42]
Make a ball of clay and embed small beads throughout it. The plum pudding model.
5 0
3 years ago
By studying fossils, scientists get an idea of ....
Tema [17]
<span>By studying fossils, scientists get an idea of what the organisms looked like, what they ate, and how they lived. </span>
6 0
3 years ago
The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe
Irina-Kira [14]

Answer:

t=147.24\ min

Explanation:

Given that:

Half life = 30 min

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30}\ min^{-1}

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration [A_0] = 7.50 mg

Final concentration [A_t] = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

0.25=7.50e^{-0.0231\times t}

750e^{-0.0231t}=25

750e^{-0.0231t}=25

x=\frac{\ln \left(30\right)}{0.0231}

t=147.24\ min

6 0
3 years ago
How much energy (heat) is required to convert 52.0 g of ice at –10.0°C to steam at 100°C? Specific heat of ice 2.09 J/g • °C Spe
dexar [7]

Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ

Explanation:

Using this formular, q = [mCpΔT] and = [nΔHfusion]

The energy that is needed in the different physical changes is thus:

The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:

q = [mCpΔT]

q = 52.0 x 2.09 x 10

q = 1.09 kJ

While from 0°C to 100°C is calculated as:

q = [mCpΔT]

q = 52.0 x 4.18 x 100

q = 21.74 kJ

And for fusion at 0°C is called Heat of fusion and would be given as:

q = n ΔHfusion

q = 52.0 / 18.02 x 6.02

q = 17.38 kJ

And that required for vaporization at 100°C is called Heat of vaporization and it's given as:

q = n ΔHvaporization

q = 52.0 / 18.02 x 40.7

q = 117.45 kJ

Add up all the energy gives 157.8 kJ

5 0
3 years ago
some components of ink are minimally attracted to the stationary phase and very soluble in the solvent. where are these componen
Trava [24]
They’ll have moved the farthest, since the solvent is best at carrying those kinds of materials.
5 0
3 years ago
Read 2 more answers
Other questions:
  • 3. How does a carbon-14 atom change back into a nitrogen atom?
    11·2 answers
  • What is the definition to squeeze a gas into a smaller space
    8·1 answer
  • When a sample of rust was mixed with acid, the rust changed into two simpler substances: W and Z. The substances W and Z could n
    15·2 answers
  • Which organelle is found in photosynthetic organisms and captures energy from the sun?
    8·2 answers
  • The atomicity of bromine is​
    11·1 answer
  • I NEED HELP IT GIVES UOU 15 POINTS
    12·1 answer
  • What state of matter has no visible shape and fills its container?
    9·2 answers
  • What element is used in bicycles?
    15·1 answer
  • Which of the following best describes asexual reproduction?
    7·2 answers
  • Match the term (Endothermic) with the correct definition
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!