Answer:
1.98 M
Explanation:
Given data
- Initial volume (V₁): 93.2 mL
- Initial concentration (C₁): 2.03 M
- Volume of water added: 3.92 L
Step 1: Convert V₁ to liters
We will use the relationship 1 L = 1000 mL.

Step 2: Calculate the final volume (V₂)
The final volume is the sum of the initial volume and the volume of water.

Step 3: Calculate the final concentration (C₂)
We will use the dilution rule.

Answer:2.86x10^-7m
Explanation:E=hc/^
E=6.94x10^-19J
c = 2.9979x10^8m/s
h= 6.626x10^-34Js
^ =( 6.626x10^-34)x( 2.9979x 10^8)/ 6.94x10^-19
= 2.86x10^-7m
<u>Answer:</u> The enthalpy of the reaction for the production of
is coming out to be -74.9 kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CH_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-74.9%29%29%5D-%5B1%5Ctimes%200%29%2B%282%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-74.9kJ)
Hence, the enthalpy of the reaction for the production of
is coming out to be -74.9 kJ
Answer:
Empirical formula is Li₂CO₃.
Explanation:
Percentage of oxygen= 65.0%
Percentage of lithium = 18.7%
Percentage of carbon= 16.3%
Empirical formula = ?
Solution:
Number of gram atoms of C = 16.3/12 = 1.4
Number of gram atoms of Li = 18.7/6.94 = 2.7
Number of gram atoms of O = 65.0/ 16 = 4.1
Atomic ratio:
Li : C : O
2.7/1.4 : 1.4/1.4 : 4.1/1.4
2 : 1 : 3
Li : C : O = 2 : 1 : 3
Empirical formula is Li₂CO₃.