Answer:
I believe it is C?
Sorry fi it doesn't help!
The liquid disilicon hexachloride reacts with water to form solid silicon dioxide, hydrogen chloride gas and hydrogen gas are
![Si{2}Cl{6} (l) + 4H{2}O (l) ------ > 2SiO{2} (s) + 6HCl (g) + H{2} (g)](https://tex.z-dn.net/?f=Si%7B2%7DCl%7B6%7D%20%28l%29%20%20%2B%20%204H%7B2%7DO%20%28l%29%20%20------%20%3E%202SiO%7B2%7D%20%28s%29%20%20%2B%20%20%20%206HCl%20%28g%29%20%20%20%20%2B%20%20%20%20%20%20H%7B2%7D%20%28g%29)
When liquid disilicon hexachloride reacts with water to form solid silicon dioxide , hydrogen chloride gas and hydrogen gas , the following chemical equation is:
![Si{2}Cl{6} (l) + H{2}O (l) ------ > SiO{2} (s) + HCl (g) + H{2} (g)](https://tex.z-dn.net/?f=Si%7B2%7DCl%7B6%7D%20%28l%29%20%20%2B%20%20H%7B2%7DO%20%28l%29%20%20------%20%3E%20SiO%7B2%7D%20%28s%29%20%20%2B%20%20%20%20HCl%20%28g%29%20%20%20%20%2B%20%20%20%20%20%20H%7B2%7D%20%28g%29)
now , we have to balance the eqaution,
Reactants product
Si 2 1
Cl 6 1
H 2 3
O 1 2
Multiply 4 with
2 with
and 6 with HCL, we get
![Si{2}Cl{6} (l) + 4H{2}O (l) ------ > 2SiO{2} (s) + 6HCl (g) + H{2} (g)](https://tex.z-dn.net/?f=Si%7B2%7DCl%7B6%7D%20%28l%29%20%20%2B%20%204H%7B2%7DO%20%28l%29%20%20------%20%3E%202SiO%7B2%7D%20%28s%29%20%20%2B%20%20%20%206HCl%20%28g%29%20%20%20%20%2B%20%20%20%20%20%20H%7B2%7D%20%28g%29)
now ,
Reactant Product
Si 2 2
Cl 6 6
H 8 8
O 4 4
Thus, we get the desire equtions.
The liquid disilicon hexachloride reacts with water to form solid silicon dioxide, hydrogen chloride gas and hydrogen gas are
![Si{2}Cl{6} (l) + 4H{2}O (l) ------ > 2SiO{2} (s) + 6HCl (g) + H{2} (g)](https://tex.z-dn.net/?f=Si%7B2%7DCl%7B6%7D%20%28l%29%20%20%2B%20%204H%7B2%7DO%20%28l%29%20%20------%20%3E%202SiO%7B2%7D%20%28s%29%20%20%2B%20%20%20%206HCl%20%28g%29%20%20%20%20%2B%20%20%20%20%20%20H%7B2%7D%20%28g%29)
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Answer:
Elemental S reacts with O2 to form SO3 according to the reaction 2S+3O2→2SO3 Part B: What is the theoretical yield of SO3 produced by the quantities described in Part A? Express your answer numerically in grams.
Part A: 1.88x10^23 O2 molecules are needed to react with 6.67 g of S.
We address the equation...
S
(
s
)
+
3
2
O
2
(
g
)
→
S
O
3
(
g
)
Explanation:
The question specifies that we got
1.88
×
10
23
dioxygen molecules
...i.e. a molar quantity of...
1.88
×
10
23
⋅
molecules
6.022
×
10
23
⋅
molecules
⋅
m
o
l
−
1
=
0.312
⋅
m
o
l
...
But we gots with respect to sulfur,
6.67
⋅
g
32.06
⋅
g
⋅
m
o
l
−
1
=
0.208
⋅
m
o
l
...
And a bit of arithmetic later, we establish that we got stoichiometric quantities of dioxygen, and sulfur….in the reaction we produce a mass of ………..
0.208
⋅
m
o
l
×
80.07
⋅
g
⋅
m
o
l
−
1
=
16.65
⋅
g
.
Note that when
sulfur trioxide
is made industrially (and this a very important commodity chemical), sulfur is oxidized to
S
O
2
, and this is then oxidized up to
S
O
3
with some catalysis...
S
O
2
(
g
)
+
1
2
O
2
(
g
)
V
2
O
5
−−→
S
O
3
(
g
)
S
O
3
(
g
)
+
H
2
O
(
l
)
→
H
2
S
O
4
(
a
q
)
sulfuric acid
The industrial sulfur cycle must be a dirty, smelly, unfriendly process. The process is undoubtedly necessary to support our civilization....
<u>Answer:</u> The number of atoms of bromine present in given number of mass is ![1.03\times 10^{22}](https://tex.z-dn.net/?f=1.03%5Ctimes%2010%5E%7B22%7D)
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Given mass of bromine = 1.37 g
Molar mass of bromine = 79.904 g/mol
Putting values in above equation, we get:
![\text{Moles of bromine}=\frac{1.37g}{79.904g/mol}=0.0171mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20bromine%7D%3D%5Cfrac%7B1.37g%7D%7B79.904g%2Fmol%7D%3D0.0171mol)
According to mole concept:
1 mole of an element contains
number of atoms.
So, 0.0171 moles of bromine will contain =
number of bromine atoms.
Hence, the number of atoms of bromine present in given number of mass is ![1.03\times 10^{22}](https://tex.z-dn.net/?f=1.03%5Ctimes%2010%5E%7B22%7D)
Bases are usually sweet and acids are sour