Explanation:
A fusion reaction is a reaction in which two or more small nuclei combine together to form a larger nuclei.
As it is known that a nucleus contains protons and neutrons. So when nuclei of two atoms combine together then there will be repulsion between the protons due to the presence of like charges. As neutrons have no charge so there will be no attraction or repulsion between neutrons of two nuclei.
Thus, we can conclude that in a fusion reaction, a major problem related to causing the nuclei to fuse into a single nucleus is the repulsion of nuclei.
Answer:
sand and water
Explanation:
when u mix these both it will become like a paste
Answer:
ΔS° = - 47.2 J/mol.K
Explanation:
ΔS°= 4(S°mH3PO4) - 6(S°mH2O) - S°mP4O10
∴ S°mH2O(l) = 69.9 J/mol.K
∴ S°mP4O10 = 231 J/mol.K
∴ S°mH3PO4 = 150.8 J/mol.K
⇒ ΔS° = 4*(150.8) - 6*(69.9) - 231
⇒ ΔS° = - 47.2 J/mol.K
Answer:
The correct statement is option c, that is, particles discharged in the air by volcanoes fall to the ground and enrich the soil.
Explanation:
The eruptions of volcanoes lead to the dispersion of ash over the broader regions surrounding the site of eruption. On the basis of the chemistry of the magma, the ash will be comprising different concentrations of soil nutrients. While the major elements found in the magma are oxygen and silica, the eruptions also lead to the discharging of carbon dioxide, water, hydrogen sulfide, sulfur dioxide, and hydrogen chloride.
In supplementation, the eruptions also discharge bits of rocks like pyroxene, potolivine, amphibole, feldspar that are in turn enriched with magnesium, iron, and potassium. As an outcome, the areas which comprise huge deposits of the volcanic soil are quite fertile.
Answer:
3. 3.45×10¯¹⁸ J.
4. 1.25×10¹⁵ Hz.
Explanation:
3. Determination of the energy of the photon.
Frequency (v) = 5.2×10¹⁵ Hz
Planck's constant (h) = 6.626×10¯³⁴ Js
Energy (E) =?
The energy of the photon can be obtained by using the following formula:
E = hv
E = 6.626×10¯³⁴ × 5.2×10¹⁵
E = 3.45×10¯¹⁸ J
Thus, the energy of the photon is 3.45×10¯¹⁸ J
4. Determination of the frequency of the radiation.
Wavelength (λ) = 2.4×10¯⁵ cm
Velocity (c) = 3×10⁸ m/s
Frequency (v) =?
Next, we shall convert 2.4×10¯⁵ cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
2.4×10¯⁵ cm = 2.4×10¯⁵ cm × 1 m /100 cm
2.4×10¯⁵ cm = 2.4×10¯⁷ m
Thus, 2.4×10¯⁵ cm is equivalent to 2.4×10¯⁷ m
Finally, we shall determine the frequency of the radiation by using the following formula as illustrated below:
Wavelength (λ) = 2.4×10¯⁷ m
Velocity (c) = 3×10⁸ m/s
Frequency (v) =?
v = c / λ
v = 3×10⁸ / 2.4×10¯⁷
v = 1.25×10¹⁵ Hz
Thus, the frequency of the radiation is 1.25×10¹⁵ Hz.
