Answer: dispersion forces, dipole-dipole and hydrogen bonds
Explanation:
Chemical reaction to biomass, like kiwi, into synthetic materials, like plastic, impact society because kiwi is biodegradable but plastic is not.
The series of chemical reactions used to change natural resources into synthetic products is termed chemical synthesis. Synthetic materials are formed when humans mix substances together for the purpose of creating new materials with desirable properties. Synthetic materials are produced by chemically changing the initial substances to create some material with different characteristics. Some of the best examples of synthetic materials are plastics and medicines. A synthetic substance may or may not be chemically identical to a naturally-occurring substance.
To know more about synthetic materials check the link below:
brainly.com/question/25834689
#SPJ4
Answer:
x(t) = −39e
−0.03t + 40.
Explanation:
Let V (t) be the volume of solution (water and
nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid
measured in liters after t minutes, and let c(t) be the concentration (by volume) of
nitric acid in solution after t minutes.
The volume of solution V (t) doesn’t change over time since the inflow and outflow
of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is
c(t) = x(t)
V (t)
=
x(t)
200
.
We model this problem as
dx
dt = I(t) − O(t),
where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,
both measured in liters of nitric acid per minute. The input rate is
I(t) = 6 Lsol.
1 min
·
20 Lnit.
100 Lsol.
=
120 Lnit.
100 min
= 1.2 Lnit./min.
The output rate is
O(t) = (6 Lsol./min)c(t) = 6 Lsol.
1 min
·
x(t) Lnit.
200 Lsol.
=
3x(t) Lnit.
100 min
= 0.03 x(t) Lnit./min.
The equation is then
dx
dt = 1.2 − 0.03x,
or
dx
dt + 0.03x = 1.2, (1)
which is a linear equation. The initial condition condition is found in the following
way:
c(0) = 0.5% = 5 Lnit.
1000 Lsol.
=
x(0) Lnit.
200 Lsol.
.
Thus x(0) = 1.
In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is
µ(t) = exp Z
P(t) dt
= exp
0.03 Z
dt
= e
0.03t
.
The solution is
x(t) = 1
µ(t)
Z
µ(t)Q(t) dt + C
= Ce−0.03t + 1.2e
−0.03t
Z
e
0.03t
dt
= Ce−0.03t +
1.2
0.03
e
−0.03t
e
0.03t
= Ce−0.03t +
1.2
0.03
= Ce−0.03t + 40.
The constant is found using x(t) = 1:
x(0) = Ce−0.03(0) + 40 = C + 40 = 1.
Thus C = −39, and the solution is
x(t) = −39e
−0.03t + 40.
Answer:
Three possible blood type alleles are Iᴬ, Iᴮ and i
Explanation:
Iᴬ, Iᴮ and i are three possible blood type alleles.
Iᴬ and Iᴮ are known as co-dominant, and The i allele is recessive.
Thus, Three possible blood type alleles are Iᴬ, Iᴮ and i
<u>-TheUnknownScientist</u>
You can detect salt in water without tasting by measuring the density of the water. Place a glass of spring water and a glass of the suspected salt water on a balance scale and the heavier one contains salt. Other ways to test for salt in water is to put a drop of water on the end of a nail and place in a gas flame. If the water contains salt, the flame will turn a yellow/orange color.
