Answer:
C6H12O6
Glucose
Explanation:
Glucose is a monocarbohydrate
carbohydrates always have only C, H, and O
brainliest plz
Answer:
CH4 + 2O2 → CO2 + 2H2O
Explanation:
This is all i could come up with im sorry.
Answer:
two examples are blood and soapy water.
Explanation:
To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
HOBr = 0.50 M
KOBr = 0.30 M = OBr-
<span> HOBr + H2O <-> H+ + OBr- </span>
<span>I 0.50 - 0 0.30 </span>
<span>C -x x x
</span>---------------------------------------------
<span>E(0.50-x) x (0.30+x) </span>
<span>Assuming that the value of x is small as compared to 0.30 and 0.50 </span>
<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>
<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48
<span>Use the van't Hoff equation:
ln
(
K2
K1
)
=
Δ
HÂş
R
(
1
T1
â’
1
T2
)
ln
(
K2
7.6*10^-3
)
=
-14,200 J
8.314
(
1
298
â’
1
333
)
ln
(
K2
7.6*10^-3
)
=
â’
1708
(
0.00035
)
ln
(
K2
0.0076
)
=
â’
0.598
Apply log rule
a
=
log
b
b
a
-0.598 =
ln
(
e
â’
0.598
)
=
ln
(
1
e
0.598
)
Multiply both sides with e^0.598
K
2
e
0.598
= 0.0076
K
e
0.598
e
0.598
=
0.0076
e
0.598
K
2
=
0.0076
e
0.598
=
4.2
â‹…
10
â’
3
K2
=
4.2
â‹…
10
â’
3</span>