<span>(a) Assuming the amount of O2(g) is not limiting the reaction, a mass of 4.23g of Ca(s) will produce an equal mass of CaO(s), hence it will produce 4.23g of CaO(s).
According to their respective molar masses, we have the following CaO molar mass :
Molar mass of Ca + Molar mass of O = 40.1 + 16 = 56.1 g/mol
4.23g of Ca will then produce : 4.23 / 56.1 = 0,07540107 mol of CaO.
(b) With the same reasonment as above, and assuming the amount of Ca is not limiting, we have :
2.87g of O2 will produce : 2.87 / 56.1 = 0,051158645 mol of CaO.
(c) From (a) and (b) answers, we can conclude that the reactant that produces less mol of CaO is limiting the reaction. Hence following the given masses, O2 is the limiting reactant.
(d) Knowing the molar mass of CaO is 56.1 g/mol and knowing that O2 is the limiting reactant, we also know the reaction can produce a maximum of 0,051158645 mol of CaO can be produced.
So we can conclude we will produce :
56.1 * 0,051158645 = 2.87g of CaO.</span>
You would place it in orbit around Earth, as most X-rays are absorbed by the atmosphere, so you would want the detector outside of the atmosphere to accurately detect the X-rays.
I think the answer is A.
A. the speed of the walkway multiplied by the speed at which she is walking
The speed she is walking accelerates her speed by the base speed of the walkway.
Answer:
4560 Torr
Explanation:
<u>Boyle's Law</u>
![\sf P_1 \cdot V_1=P_2 \cdot V_2](https://tex.z-dn.net/?f=%5Csf%20P_1%20%5Ccdot%20V_1%3DP_2%20%5Ccdot%20V_2)
where:
= initial volume
= final volume
= initial pressure
= final pressure
Given:
Substituting the given values into the formula:
![\implies \sf 1.5 \cdot 4=P_2 \cdot 1](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%201.5%20%5Ccdot%204%3DP_2%20%5Ccdot%201)
![\implies \sf P_2=6\:atm](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20P_2%3D6%5C%3Aatm)
To convert atm to Torr, multiply atm by 760:
![\implies \sf P_2=6\cdot 760=4560\:Torr](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20P_2%3D6%5Ccdot%20760%3D4560%5C%3ATorr)
A pond of cement is more dense