Answer:
Amplitude does not effect the wavelength in linear system.
Explanation:
Amplitude:
It is the measure of height from peak to trough.
Wavelength:
It is measure of length from peak to peak.
There is no relation between the amplitude and wavelength but if the system will no more linear then high amplitude can cause the distortion in wave if more frequencies are present. However frequency and wavelength can be related. The wave with higher frequency have shorter wavelength and vise versa.
Frequency:
It is the number of waves passing through a given point in a given time period. It is measure in Hz or s⁻¹
Answer:
The products are carbon dioxide and water
Explanation:
Step 1: Data given
Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
Step 2: The complete combustion of C3H7OH:
For the combustion of 1-propanol, we need O2.
The products of this combustion are CO2 and H2O.
C3H7OH + O2→ CO2 + H2O
On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3
C3H7OH + O2→ 3CO2 + H2O
On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.
C3H7OH + O2→ 3CO2 + 4H2O
On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).
To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.
2C3H7OH + 9O2→ 6CO2 + 8H2O
For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O
The products are carbon dioxide and water
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
Answer:
1)alkali metals
2)1g/ml
3) mercury
Explanation:
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