This problem requires our calculation to undergo the dimensional analysis approach. In this approach, you disregard the actual quantity and focus on the units of measurement. This helps us know the units of our final answer.
First, let's ignore 16. Let's focus on converting the units kPa-mm³/s to mJ/s. The unit kPa stands for kiloPascals which is 1000 times greater than 1 Pa. The unit mJ, on the other hand, stands for millijoules, which is 1000 times lesser than Joules. The relationship between the two is that, Joules = Pa × m³. But since we want our final answer to be mJ, that would be equal to Pa×mm³. Since the original unit already contains mm³, all we have to do is convert kPa to Pa.
16 kPa-mm³/s * (1000 Pa/1 kPa) = 16,000 Pa-mm³/s
Since Pa-mm³ is equal to mJ, the final conversion yields to 16,000 Pa-mm³/s.
Answer:
Explanation:
Given
Required
Convert to meters per minutes
Start by converting the speed from miles to meters
So, we have:
Next, we convert time from hours to minutes
So, we have:
<em>Hence, the equivalent of 25 miles per hour is 670.56 meters per minutes</em>
Answer:
The new element will be thorium-226 (²²⁶Th).
Explanation:
The beta decay is given by:
Where:
A: is the mass number
Z: is the number of protons
β⁻: is a beta particle = electron
: is an antineutrino
The neutral atom has 88 electrons, so:
Hence the element is radium (Ra), it has A = 226.
If Ra undergoes 2 rounds of beta minus decay, we have:
Therefore, if a neutral atom with 88 electrons undergoes 2 rounds of beta minus decay the new element will be thorium-226 (²²⁶Th).
I hope it helps you!
Answer:
It should be Magnetic fields can push and pull objects without touching
but I only say with touching. Maybe it's an error on you're teachers/Prof behalf.
Answer:
Explanation:
![\Delta H_{rxn}^{0}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5Csum%20%5Bn_%7Bi%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28product%29_%7Bi%7D%5D-%5Csum%20%5Bn_%7Bj%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28reactant_%7Bj%7D%29%5D)
Where 
and 
are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).
is standard heat of formation and 
is standard enthalpy change for reaction at 
So, ![\Delta H_{rxn}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
or, ![\Delta H_{rxn}=[3mol\times -393.509kJ/mol]+[4mol\times -241.818kJ/mol]-[1mol\times -103.8kJ/mol]-[5mol\times 0kJ/mol]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B3mol%5Ctimes%20-393.509kJ%2Fmol%5D%2B%5B4mol%5Ctimes%20-241.818kJ%2Fmol%5D-%5B1mol%5Ctimes%20-103.8kJ%2Fmol%5D-%5B5mol%5Ctimes%200kJ%2Fmol%5D)
or,
