Answer:
frequency = 0.47×10⁴ Hz
Explanation:
Given data:
Wavelength of wave = 6.4× 10⁴ m
Frequency of wave = ?
Solution:
Formula:
Speed of wave = wavelength × frequency
Speed of wave = 3 × 10⁸ m/s
Now we will put the values in formula.
3 × 10⁸ m/s = 6.4× 10⁴ m × frequency
frequency = 3 × 10⁸ m/s / 6.4× 10⁴ m
frequency = 0.47×10⁴ /s
s⁻¹ = Hz
frequency = 0.47×10⁴ Hz
Thus the wave with wavelength of 6.4× 10⁴ m have 0.47×10⁴ Hz frequency.
Answer:
Explanation:
2S + 3O₂ = 2SO₃
2moles 3 moles
2 moles of S react with 3 moles of O₂
5 moles of S will react with 3 x 5 / 2 moles of O₂
= 7.5 moles of O₂ .
O₂ remaining unreacted = 10 - 7.5 = 2.5 moles .
All the moles of S will exhausted in the reaction and 2.5 moles of oxygen will be left .
Answer:
The correct answer is - C: Energy is neither created nor destroyed, but continually transformed into new forms of energy.
Explanation:
Conservation of energy is the first law of thermodynamics that says that energy is neither generated nor destroyed, can only be transformed from one form to another form continuously. That means no matter what the condition the amount of energy will remain the same, only form will be changed.
For instance, if an electric current will pass through a circuit of a light bulb the electrical energy present in the circuit will transfer into heat and light energy, The amount will remain constant but the one form change to another.
Answer:
dipole-dipole forces, ion-dipole forces, higher molar mass, hydrogen bonding, stronger intermolecular forces
Explanation:
<em>1. H₂S and H₂Se exhibit the following intermolecular forces: </em><em>dipole-dipole forces </em><em>and </em><em>ion-dipole forces</em><em>.</em> These molecules have a bent geometry, thus, a dipolar moment which makes them dipoles. When they are in the aqueous form they are weak electrolytes whose ions interact with the water dipoles
<em>2. Therefore, when comparing H₂S and H₂Se the one with a </em><em>higher molar mass</em><em> has a higher boiling point.</em> In this case, H₂Se has a higher boiling point than H₂S due to its higher molar mass.
<em>3. The strongest intermolecular force exhibited by H₂O is </em><em>hydrogen bonding</em><em>. </em>This is a specially strong dipole-dipole interaction in which the positive density charge on the hydrogens is attracted to the negative density charge on the oxygen.
<em>4. Therefore, when comparing H₂Se and H₂O the one with </em><em>stronger intermolecular forces</em><em> has a higher boiling point. </em>That's why the boiling point of H₂O is much higher than the boiling point of H₂Se.
Answer:
An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.
The answer to the above question is
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
Explanation:
To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows
ΔH = m×c×ΔT
= 0.205×4190×(79.9 -31.0) = 42002.655 J
Therefore fore the ice, we have
Total heat = mi×L + mi×ci×ΔTi = mi×334000 + mi × 2100 × (0 -−17.5) = 42002.655 J
370750×mi = 42002.655 J
or mi = 0.1133 kg
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
