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lidiya [134]
2 years ago
8

Select three (3) types of waves that can be harmful to life.

Chemistry
2 answers:
pychu [463]2 years ago
7 0

Answer:

opt a f g is correct answer ok

zubka84 [21]2 years ago
7 0

Answer:

x-rays,gamma rays and ultraviolet rays are harmful to life.

Explanation:

hope this will hep you!

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5. What is the mole fraction of NaOH in an aqueous<br> solution that contains 31.0 % NaOH by mass?
Liono4ka [1.6K]

Answer:

16.8%      

Explanation:

31% NaOH    molar mass  40 gm

69% H2O      molar mass  18  gm

1000 gm would be  

 310 gm NaOH  or  310/40 = 7.75 moles

 690 gm of H2O or  690/18 = 38.333 moles

7.75 / (7.75 + 38.333) = .168   mole fraction

6 0
2 years ago
How do you write the balanced equation ​
katrin [286]

Answer:

See BELOW!

Explanation:

Your question is confusing, you are not mentioning specifically what you are trying to balance the equation of. If you require more assistance, write in the comments and I'll be glad to assist. you!

6 0
3 years ago
Please help me ASAP!!!
DENIUS [597]

Answer: Patrick was so sure he was a thief because why would you knock if it was your own room you would open the door. And he didn’t have a key to open the door.

Hope this helped

Explanation:

3 0
3 years ago
Which of the following is a transuranium element?<br> Ra<br> Am<br> Tc<br> Pa
lyudmila [28]
Am - it has an atomic number of 95 which is greater than 92.

Transuranium elements are elements with atomic levels greater than 92
3 0
3 years ago
For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the v
Ierofanga [76]

Answer:

\Delta G^o=-5.4032 kJ

The temperature for \Delta G^o=0[/tex is [tex]T=328.6 K

Explanation:

The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:

\Delta G^o=\Delta H^o + T*\Delta S^o

Where:

\Delta G^o is Gibbs's energy in kJ

\Delta H^o is the enthalpy in kJ

\Delta S^o is the entropy in kJ/K

T is the temperature in K

Solving:

\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K

\Delta G^o=-5.4032 kJ

For \Delta G^o=0:

0=\Delta H^o - T*\Delta S^o

\Delta H^o= T*\Delta S^o

T=\frac{\Delta H^o}{\Delta S^o}

T=\frac{-58.03 kJ}{-0.1766 kJ/K}

T=328.6 K

3 0
3 years ago
Read 2 more answers
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