Answer:
The answer to your question is 245 grams
Explanation:
Data
Volume 6.5 L
Molarity = 0.34
mass of CaCl₂ = ?
Process
1.- Calculate the molar mass of CaCl₂
molar mass = (1 x 40) + (2 x 35.5)
= 40 + 71
= 111 g
2.- Convert the grams to moles
111 g of CaCl₂ -------------- 1 mol
x ---------------0.34 mol
x = (0.34 x 111) / 1
x = 37.74 g
3.- Calculate the total mass
37.74 g ------------------ 1 L
x ------------------ 6.5 L
x = (6.5 x 37.74) / 1
x = 245.31
<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:
where,
= half life of the reaction = 5730 years
Putting values in above equation, we get:
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = ? yr
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the sample = 100 grams
[A] = amount left after decay process = (100 - 25) = 75 grams
Putting values in above equation, we get:
Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %
Answer:
5 moles of NO₂ will remain after the reaction is complete
Explanation:
We state the reaction:
3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g)
3 moles of nitric oxide can react with 1 mol of water. Ratio is 3:1, so we make this rule of three:
If 3 moles of nitric oxide need 1 mol of water to react
Then, 26 moles of NO₂ may need (26 .1) / 3 = 8.67 moles of H₂O
We have 7 moles of water but we need 8.67 moles, so water is the limiting reactant because we do not have enough. In conclusion, the oxide is the reagent in excess. We can verify:
1 mol of water needs 3 moles of oxide to react
Therefore, 7 moles of water will need (7 .3)/1 = 21 moles of oxide
We have 26 moles of NO₂ and we need 21, so we still have oxide after the reaction is complete. We will have (26-21) = 5 moles of oxide that remains
<span>1.05 g/ml * 1000 ml = 1050g/l because of 1g/ml = 1 kg/l
so, a/q
mass of 4.7 l of whole blood in pound =
4.7 * 1050 = 4935 g
so in pound
4935g = 10.87981p</span>
Answer: Enthalpy of combustion (per mole) of 
is -2657.5 kJ
Explanation:
The chemical equation for the combustion of butane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20%5CDelta%20H%5Eo_f_%7BCO_2%28g%29%7D%29%2B%2810%5Ctimes%20%5CDelta%20H%5Eo_f_%7BH_2O%28g%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7BC_4H_%7B10%7D%28g%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7BO_2%28g%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.6%29%2B%284%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-5315kJ)
Enthalpy of combustion (per mole) of is -2657.5 kJ
