First off, With a hypothesis
Explanation:
The transuranium elements are produced by the capture of neutrons
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Answer:
Water's polarity and its ability to form hydrogen bonds give it special properties, including adhesion, cohesion, and the ability to moderate temperature.
We are given an equation 2Mg+O2-> 2MgO and a starting chemical Mg of about 212 g. In order to solve for the amount of O2 needed, we need the molecular weight of Mg and O2.
Molecular weight:
Mg=24.305 g/mol
O2=16(2)=32 g/mol
Note that for every 1 mol of O2, the amount of Mg must be 2 mol.
So,
g O2 = 212 g Mg x1mo Mgl/24.305 g Mg x1mol O2 /2 mol Mg x 32 g O2/mol O2
gO2=139.56 g
Therefore, 139.56 g of O2 is needed for every 212 g Mg.
Answer:
0.41kg/sec
Explanation:
PV= nRT
Given : V= 505 L
P=0.88 atm
R= 0.08206 Latm/K*mol
T= 172 .0C = 172+273 = 445 K
n = PV /RT = 0.88 * 505 / 0.08206 * 445 = 12.17 moles per sec of N2 are consumed
As per reaction : N2 + 3H2 ----> 2NH3
1 mole N2 is consumed to produce 2 moles NH3
moles of NH3 produced per sec :
(2 moles NH3/1mol N2) * 12.17 moles N2 = 24.34 moles NH3 per sec
grams of NH3 produced per sec =
24.34 moles NH3 per sec * molar mass NH3 = 24.34 moles NH3 per sec * 17.031 g/mol = 414.5 g NH3 per sec
rate in Kg/sec = 414.5 g NH3 per sec * (1kg /1000g) = 0.4145 Kg/sec
= 0.41kg/sec
