Answer:
The sum of positive and negative charges in a unit of Al2O3 equals zero.
Aluminium has a charge of +3 while Oxygen has a charge of -2 on each ion.
Al203 has 2 Al atoms and 3 O atoms.
Charge on Al2O3 = 2(charge on Al ion) + 3(charge on O ion)
= 2(3) + 3(-2)
= 6 - 6
= 0
Explanation:
Aluminium has 3 electrons in the outermost shell and has the tendency to lose those 3 electrons to form a positive ion and have a complete outermost shell.
Whereas, Oxygen has 6 electrons in the outermost and has the tendency to accept two more electrons to form a negative ion and have a complete outermost shell.
Answer:
1, 2, and 3.
Explanation:
Hello.
In this process, since the phase transitions that require energy are those that pass from a state with less energy or more molecular order to a state with more energy or less molecular order, say, from solid to liquid (melting), from liquid to gas (boiling) and from solid to gas (sublimation), we can conclude that the arrows representing heat energy gained are 1, 2, and 3 since 1 represents boiling, 2 melting and 3 sublimation.
Best regards.
It's cold outside, the water vaper in your breath condenses into tiny droplets of liquid water and ice that you can see.
Answer:
Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).:
Answer:
solution:
to find the speed of a jogger use the following relation:
V
=
d
x
/d
t
=
7.5
×m
i
/
h
r
...........................(
1
)
in Above equation in x and t. Separating the variables and integrating,
∫
d
x
/7.5
×=
∫
d
t
+
C
or
−
4.7619
=
t
+
C
Here C =constant of integration.
x
=
0 at t
=
0
, we get: C
=
−
4.7619
now we have the relation to find the position and time for the jogger as:
−
4.7619 =
t
−
4.7619
.
.
.
.
.
.
.
.
.
(
2
)
Here
x is measured in miles and t in hours.
(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),
to get:
= −
4.7619
=
1
−
4.7619
= −
3.7619
or x
=
7.15
m
i
l
e
s
(b) To find the jogger's acceleration in m
i
l
/
differentiate
equation (1) with respect to time.
we have to eliminate x from the equation (1) using equation (2).
Eliminating x we get:
v
=
7.5×
Now differentiating above equation w.r.t time we get:
a
=
d
v/
d
t
=
−
0.675
/
At
t
=
0
the joggers acceleration is :
a
=
−
0.675
m
i
l
/
=
−
4.34
×
f
t
/
(c) required time for the jogger to run 6 miles is obtained by setting
x
=
6 in equation (2). We get:
−
4.7619
(
1
−
(
0.04
×
6 )
)^
7
/
10=
t
−
4.7619
or
t
=
0.832
h
r
s