Answer:
Initial concentration of the reactant = 3.34 × 10^(-2)M
Explanation:
Rate of reaction = 2.30×10−4 M/s,
Time of reaction = 80s
Final concentration = 1.50×10−2 M
Initial concentration = Rate of reaction × Time of reaction + Final concentration
= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M
Initial concentration = 3.34 × 10^(-2)M
Answer:
vector quantities are resolved into their component form (along the x and y-axis) before adding them. Let us assume that two vectors are
→
a
=
x
1
^
i
+
y
1
^
j
and
→
b
=
x
2
^
i
+
y
2
^
j
, we can find the sum of two vectors as follows.
→
a
+
→
b
=
x
1
^
i
+
y
1
^
j
+
x
2
^
i
+
y
2
^
j
=
(
x
1
+
x
2
)
^
i
+
(
y
1
+
y
2
)
^
j
The direction of the sum of the vectors (with positive x-axis) is,
θ
=
tan
−
1
(
y
1
+
y
2
x
1
+
x
2
)
Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
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from one energy form to one
