Answer: -227 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%20n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BC_2H_2%7D%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%282%5Ctimes%20-393.5%29%2B%281%5Ctimes%20-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28%5Cfrac%7B5%7D%7B2%7D%5Ctimes%200%29%5D)
![-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%28-787%29%2B%28-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%280%29%5D)
Therefore, the enthalpy change for 
is -227 kJ.
Answer:
the ball will move towards the big bully
Answer:
0.521 moles still present in the container.
Explanation:
It is possible to answer this question by using the general gas law, that is:
PV = nRT
<em>Where P represents pressure of the gas, v its volume, n moles, R gas constant law and T absolute temperature (21.7°C + 273.15 = 294.85K)</em>
Replacing with values of the initial conditions of the container, its volume is:
V = nRT / P
V = 2.00mol*0.082atmL/molK*294.85K / 3.75atm
V = 12.9L
When some gas is released, absolute temperature is 28.1°C + 273.15 = 301.25K, the pressure is 0.998atm and <em>the volume of the container still constant. </em>Again, using general gas law:
PV / RT = n
0.998atm*12.9L / 0.082atmL/molK*301.25K = n
0.521 moles = n
<h3>0.521 moles still present in the container.</h3>
<em />
Molarity = number of mole of substance(n) / volume of solution (V).
n(CaCl2) = mass (CaCl2)/M(CaCl2)
M(CaCl2) = 40+2*35.5 = 111 g/mol
n(CaCl2) =39.5 g CaCl2*1 mol/111g
0.250 M = 39.5 g CaCl2*1 mol/111g*volume of solution (V).
volume of solution (V) = 39.5 g CaCl2*1 mol/(0.250 M*111g) = 1.42 L
Answer : Total molecules that will be needed to visualize a single egg will be 78500 molecules of dye.
Explanation : As a single egg cell has an approximately diameter of 100 μm.
We can use this formula to calculate area of the cell membrane;
A = π 
;
We can take π as 3.14 and we get;
A = 3.14 X
Soving we get;
A = 7850 μ
Here we have to calculate the amount of dye molecules which will be needed for 10 fluorescent molecules / μ but;
here 1 μ = 7850 μ dye molecules.
Therefore, 10 fluorescent molecules will need;
7850 X 10 = 78500 molecules of dye.
Therefore, the answer is 78500 molecules of dye.
