Full Question:
A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?
2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)
Answer:
13.1 g K2CO3 required to neutralize spill
Explanation:
2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)
Number of moles = Volume * Molar Concentration
moles HBr= 0.42L x .45 M= 0.189 moles HBr
From the stoichiometry of the reaction;
1 mole of K2CO3 reacts with 2 moles of HBr
1 mole = 2 mole
x mole = 0.189
x = 0.189 / 2 = 0.0945 moles
Mass = Number of moles * Molar mass
Mass = 0.0945 * 138.21 = 13.1 g
Answer:
1.08 M
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 45 mL
Initial concentration (C₁) = 6 M
Final volume (V₂) = 250 mL
Final concentration (C₂) =?
The final concentration of the solution can be obtained by using the dilution formula as illustrated below:
C₁V₁ = C₂V₂
6 × 45 = C₂ × 250
270 = C₂ × 250
Divide both side by 250
C₂ = 270 / 250
C₂ = 1.08 M
Therefore, the final concentration of the solution is 1.08 M.
Idk why but for some reason the picture isnt ahowing up
Answer:
at 181.0 
is -723.3 kJ/mol.
Explanation:
We know, 
where, T is temperature in kelvin.
Let's assume 
and 
does not change in the temperature range 25.0 - 181.0 .
= (273+181.0) K = 454.0 K
Hence, at 181.0 , ![\Delta G^{0}=(-795.8kJ/mol)-[(454.0 K)\times (-159.8\times 10^{-3}kJ/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20G%5E%7B0%7D%3D%28-795.8kJ%2Fmol%29-%5B%28454.0%20K%29%5Ctimes%20%28-159.8%5Ctimes%2010%5E%7B-3%7DkJ%2FK.mol%29%5D)
= -723.3 kJ/mol
