Answer:
D) With an increase in altitude, atmospheric pressure increases as well.
Explanation:
Generally when altitude increases, the value of pressure decreases. This shows that pressure is inversely proportional to altitude. For example, the higher the altitude, the lower the pressure and vice versa. At very high altitude, the number of molecules of air are smaller than the number of moles of air at very low altitude. Thus, the higher the altitude, the lower the atmospheric pressure and the lower the altitude, the higher the atmospheric pressure. Therefore, option (D) is false.
<h3>
Answer:</h3>
0.387 J/g°C
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Explanation:</h3>
- To calculate the amount of heat absorbed or released by a substance we need to know its mass, change in temperature and its specific heat capacity.
- Then to get quantity of heat absorbed or lost we multiply mass by specific heat capacity and change in temperature.
- That is, Q = mcΔT
in our question we are given;
Mass of copper, m as 95.4 g
Initial temperature = 25 °C
Final temperature = 48 °C
Thus, change in temperature, ΔT = 23°C
Quantity of heat absorbed, Q as 849 J
We are required to calculate the specific heat capacity of copper
Rearranging the formula we get
c = Q ÷ mΔT
Therefore,
Specific heat capacity, c = 849 J ÷ (95.4 g × 23°C)
= 0.3869 J/g°C
= 0.387 J/g°C
Therefore, the specific heat capacity of copper is 0.387 J/g°C
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
Answer:
E = 29.7× 10⁻²⁰ j
Explanation:
Given data;
Frequency of light = 4.48 × 10¹⁴ Hz
Energy of photon = ?
Solution:
Formula:
E = h.f
E = energy of photon
h = planck's constant
f = frequency
E = h.f
E = 6.63 × 10⁻³⁴ Kg.m² /s × 4.48 × 10¹⁴ s⁻¹
E = 29.7× 10⁻²⁰ Kg.m²/s²
Kg.m²/s² = j
E = 29.7× 10⁻²⁰ j
