Answer:
4.84 × 10⁻⁶ M
Explanation:
First, we will calculate the partial pressure of Ar (pAr) using the following expression.
pAr = P × χAr
where,
P: total pressure
χAr: mole fraction
pAr = P × χAr
pAr = 0.370 atm × 9.34 × 10⁻³
pAr = 3.46 × 10⁻³ atm
We can find the solubility of Ar in water (S) using Henry's law.
S = kH × pAr
where
kH: Henry's constant
S = kH × pAr
S = 1.40 × 10⁻³ M/atm × 3.46 × 10⁻³ atm
S = 4.84 × 10⁻⁶ M
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
Answer:
imp not sure but i think the density is 5
Explanation:
im probably wrong but i think this because the water rose to the 50 ml mark when before it was 45 then again i am only in 8th grade but it makes sense to me
Answer: The main purpose of any warhead is to inflict damage on the target. The way the damage is caused may vary with different types of warheads, but in the most general sense, damage is caused by the transfer of energy from the warhead to the target. The energy is typically mechanical in nature and takes the form of a shock wave or the kinetic energy of fragments. In either case, a large amount of energy must be released. For many warheads that energy is stored in the form of chemical explosives.
Explosive Reactions
There are many chemical reactions that will release energy. These are known as exothermic reactions. If the reaction proceeds slowly, the released energy will be dissipated and there will be few noticeable effects other than an increase in temperature. On the other hand, if the reaction proceeds very rapidly, then the energy will not be dissipated. Thus, a great quantity of energy can be deposited into a relatively small volume, then manifest itself by a rapid expansion of hot gases, which in turn can create a shock wave or propel fragments outwards at high speed. Chemical explosions may be distinguished from other exothermic reactions by the extreme rapidity of their reactions. In addition to the violent release of energy, chemical explosions must provide a means to transfer the energy into mechanical work. This is accomplished by expanding product gases from the reaction. If no gases are produced, then the energy will remain in the products as heat.
Most chemical explosions involve a limited set of simple reactions, all of which involve oxidation (reaction with oxygen). A relatively easy way to balance chemical explosive equations is to assume that the following partial reactions take place to their maximum extent (meaning one of the reactants is totally consumed) and in order of precedence:
Explanation:
Answer:
D
Explanation:
It was later identified that atoms of the same element can be different. This is mostly seen in elements with different isotopes. An example is carbon-14 and carbon-12 that have different masses due to differences in neutrons numbers in their nuclei.
Atoms are also divisible into subatomic particles. Today, atoms can be smashed apart into neutrons, protons and electrons particles. This also occurs naturally in radioactive decay.
