The answer is 98ppm.
The ppm (Parts per million) is also a concentration unit. 1 ppm is equivalent to 1mg/L
Now, concentration of the solution of sulfuric acid is 0.0980 g/L.
We rewrite, 0.0980 g = 0.0980*1000 mg = 98mg
Therefore, the concentration of the sulfuric acid solution is 98 mg/L = 98 ppm.
i think its C pls dont get mad if it is wrong
For the absorbance of the solution in a 1.00 cm cell at 500 nm is mathematically given as
A’ = 0.16138
<h3>What is the absorbance of the solution in a 1.00 cm cell at 500 nm?</h3>
Absorbance (A) 2 – log (%T) = 2 – log (15.6) = 0.8069
Generally, the equation for the Beer’s law is mathematically given as
A = ε*c*l
0.8069 = ε*c*(5.00 )
ε*c = 0.16138 cm-1
then for when ε*c is constant
l’ = 1.00
A’ = (0.16138 cm-1)*(1.00 cm)
A’ = 0.16138
In conclusion, the absorbance of the solution in a 1.00 cm cell at 500 nm is
A’ = 0.16138
Read more about Wavelength
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The balanced chemical reaction for the complete combustion of C4H10 is shown below:
C4H10 + (3/2)O2 --> 4CO2 + 5H2O
The enthalpy of formation are listed below:
C4H10: -2876.9 kJ/mol
O2: none (because it is pure substance)
CO2: -393.5 kJ/mol
H2O: -285.8 kJ/mol
The enthalpy of combustion is computed by subtracting the total enthalpy formation of the reactants from that of the products.
ΔHc = (4)(-393.5 kJ/mol) + (5)(-285.8 kJ/mol) - (-2876.9 kJ/mol)
= -<em>126.1 kJ</em>
Thus, the enthalpy of combustion of the carbon is -126.1 kJ.
In this compound (Phosgene) the central atom (carbon is Sp² Hybridized).
Sp, Sp² and Sp³ can be calculated very simply by doing three steps,
Step 1:
Assume triple bond and double bond as one bond and assign s or p to it. In this example carbon double bond oxygen is considered once and let suppose it is s. Now we are having our s.
Step 2:
Count lone pair of electron, each lone pair counts for s and p. In this case there is no lone pair of electron on carbon, so not included.
Step 3:
Count single bonds for s and p. As we have already assigned s to the double bond, now one p for one single bond, and other p for the other single bond.
Result:
So, we counted 1 s for double bond, 1 p for one single and other p for second single bond. As a whole we got,
Sp²
Practice:
You can practice for hybridization of Oxygen in this molecule. Oxygen has 2 lone pair of electrons. (Hint: Sp² Hybridization)
