<span>I think it is the barometer and aneroid barometer</span>
Carbon dioxide is a compound. Oxygen is an element. A compound is 2 or more elements chemically combined. An element is a pure substance. A mixture is combining two or more different compounds/elements.
Answer:
Explanation:
CH₃Br+NaOH⟶CH₃OH+NaBr
It is a single step bimolecular reaction so order of reaction is 2 , one for CH₃Br and one for NaOH .
rate of reaction = k x [CH₃Br] [ NaOH]
.008 = k x .12 x .12
k = .55555
when concentration of CH₃Br is doubled
rate of reaction = .555555 x [.24] [ .12 ]
= .016 M/s
when concentration of NaOH is halved
rate of reaction = .555555 x [.12] [ .06 ]
= .004 M/s
when concentration of both CH₃Br and Na OH is made 5 times
rate of reaction = .555555 x .6 x .6
= 0.2 M/s
Answer:
I am explain you in image
4.20 mol Al would react completely with 4.20 x (1/2) = 2.10 mol Fe2O3, but there is not that much Fe2O3 present, so Fe2O3 is the limiting reactant. (1.75 mol Fe2O3) x (2/1) x ( 55.8452 g Fe/mol) = 195 g Fe 3 MgO + 2 H3PO4 → Mg3(PO4)2 + 3 H2O (15.0 g MgO) / (40.3045 g MgO/mol) = 0.37217 mol MgO (18.5 g H3PO4) / (97.9953 g H3PO4/mol) = 0.18878 mol H3PO4 0.18878 mol H3PO4 would react completely with 0.18878 x (3/2) = 0.28317 mole of MgO, but there is more MgO present than that, so MgO is in excess and H3PO4 is the limiting reactant. Now we must consider why the problem tells us "17.6g of Mg3(PO4)2 is obtained". The first possibility is that it's just there for the sake of confusion -- in which case ignore it and proceed this way: ((0.37217 mol MgO initially) - (0.28317 mole MgO reacted)) x (40.3045 g MgO/mol) = 3.59 g MgO left over However, if the amount of magnesium phosphate obtained is given because the reaction was stopped before it was complete, the amount obtained governs the amount reacted and the amount left over, so proceed this way: (17.6g Mg3(PO4)2) / (262.8581 g Mg3(PO4)2/mol) x (3/1) = 0.20087 mol MgO reacted ((0.37217 mol MgO initially) - (0.20087 mole MgO reacted)) x (40.3045 g MgO/mol) = 6.90 MgO left over
