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Marysya12 [62]
3 years ago
5

A thermogram is an image produced

Chemistry
1 answer:
Y_Kistochka [10]3 years ago
4 0

Answer: C

infrared rays

Explanation:

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How mant atom arre in this compound: 2H2O
Karolina [17]

Answer:

3 atoms

Explanation:

it contains 2 atoms of hydrogen(H) and 1 atom of oxygen(O). So in total there are 3 atoms.

6 0
3 years ago
which has more kinetic energy, a 5-kg object moving at 5 m/s 1 m off the ground or a 5-kg object at rest 2 m off the ground?
irina1246 [14]

Answer:

Explanation:

Kinetic energy is the energy due to changes in position of a body. It is always with regards to motion of a body.

Potential energy on the other hand is the energy at rest of a body.

To estimate kinetic energy, we use the formula:

      K.E =  \frac{1}{2}m v^{2}

where m is the mass of the body and v is the velocity of the object.

mass is 5kg and velocity is 5ms⁻¹

      K.E =  \frac{1}{2}5 x  5^{2}

      K.E = 62.5J

for the object at rest;

    potential energy is calculated:

    P.E = mgh

where m is the mass, g is the acceleration due to gravity and h is the height

m is 5kg, h is 2 and g is 9.8

    P.E = 5 x 9.8 x 2 = 98J

The moving object has kinetic energy with the object  at rest having potential energy.

4 0
3 years ago
The reactive properties or chemical behavior of an atom mostly depend on the number of
Kryger [21]
The reactive properties or chemical behavior of an atom is mostly dependant on the number of electrons in the outer shell
7 0
3 years ago
4. Write the chemical equation for the reaction that occurs when baking soda (NaHCO3) is
aev [14]

Answer: HCl+NaHCO₃=NaCl+CO₂+H₂O

Explanation:

3 0
3 years ago
If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series em
erica [24]

Answer: Wavelength  associated with the fifth line is 397 nm

Explanation:

E=\frac{hc}{\lambda}

\lambda = Wavelength of radiation

E= energy

For fifth line in the H atom spectrum in the balmer series will be from n= 2 to n=7.

Using Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant =10973731.6m^{-1}

n_f = Higher energy level = 7

n_i= Lower energy level = 2 (Balmer series)

Putting the values, in above equation, we get

\frac{1}{\lambda}=10973731.6\times \left(\frac{1}{2^2}-\frac{1}{7^2} \right )\times 1^2

\frac{1}{\lambda}=2.52\times 10^{6}m

\lambda}=3.97\times 10^{-7}m=397 nm      1nm=10^{-9}m

Thus wavelength λ associated with the fifth line is 397 nm

7 0
4 years ago
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