45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)
The balanced equation for the reaction is given below:
2Al + 3Cl₂ —> 2AlCl₃
From the balanced equation above,
2 atoms of Al required 3 molecules of Cl₂.
With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:
From the balanced equation above,
2 atoms of Al required 3 molecules of Cl₂.
Therefore,
30 atoms of Al will require = 
= 45 molecules of Cl₂.
Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)
Learn more: brainly.com/question/24918379
For this, we first calculate molecular weight of MgSiO₃:
Atomic masses:
Mg = 24
Si = 28
O = 16
Mr = 24 + 28 + 16 x 3
Mr = 100
moles = mass / Mr
moles = 237 / 100
moles = 2.37
Answer:
0.5 moles of LiOH will absorb 5.6 L of 
Explanation:
According to law of conservation of mass, the sum of mass on the reactant side must be equal to the sum of mass on product side.
The balanced chemical equation is:
2 moles of LiOH absorb 1 mole of i.e. 22.4 Liters at STP
0.5 moles of LiOH absorb
=
0.5 moles of LiOH will absorb 5.6 L of
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
