Answer:
Alkali metal hydroxides can be used to test the identity of metals in certain salts. The colour of the precipitate will help identify the metal : Calcium hydroxide is soluble; no precipitate is formed.
Answer:
22 in total but round it up to 23
1 CH4 (g) + 2 O2 (g) -----> CO2 (g) + 2H2O(l) ΔH= - 890 kJ
1 mol 2 mol
1) If ΔH has minus, it means "release". We need only "release" choices.<span>
2) From reaction
1 mol </span>CH4 (g) "releases" ΔH= - 890 kJ - We do not have this choice.
2 mol O2 (g) "release" ΔH= - 890 kJ, so
1 mol O2 (g) "release" ΔH= - 445 kJ
Correct answer is B.
We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
V₁T₂ = V₂T₁
Substituting the known values,
(0.456 L)(65 + 273.15) = (3.4 L)(T₁)
T₁ = 45.33 K
Answer:
6.52×10⁴ GHz
Explanation:
From the question given above, the following data were obtained:
Wavelength (λ) = 4.6 μm
Velocity of light (v) = 2.998×10⁸ m/s
Frequency (f) =?
Next we shall convert 4.6 μm to metre (m). This can be obtained as follow:
1 μm = 1×10¯⁶ m
Therefore,
4.6 μm = 4.6 μm × 1×10¯⁶ m / 1 μm
4.6 μm = 4.6×10¯⁶ m
Next, we shall determine frequency of the light. This can be obtained as follow:
Wavelength (λ) = 4.6×10¯⁶ m
Velocity of light (v) = 2.998×10⁸ m/s
Frequency (f) =?
v = λf
2.998×10⁸ = 4.6×10¯⁶ × f
Divide both side by 4.6×10¯⁶
f = 2.998×10⁸ / 4.6×10¯⁶
f = 6.52×10¹³ Hz
Finally, we shall convert 6.52×10¹³ Hz to gigahertz. This can be obtained as follow:
1 Hz = 1×10¯⁹ GHz
Therefore,
6.52×10¹³ Hz = 6.52×10¹³ Hz × 1×10¯⁹ GHz / 1Hz
6.52×10¹³ Hz = 6.52×10⁴ GHz
Thus, the frequency of the light is 6.52×10⁴ GHz
