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icang [17]
3 years ago
14

How does the volume occupied by a cubic centimeter compare with the volume occupied by a millimeter

Chemistry
1 answer:
Bad White [126]3 years ago
7 0
I'm pretty sure it's the same 2mm = 2cc = 2c^2
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The number of electrons in an oxide ion O^2- is?
Tatiana [17]

10 electrons are in o^2

7 0
3 years ago
P 1. Elements X and Y react to form an ionic compound a formula XY. What could be the proton (atomic) numbers of X and Y?
larisa86 [58]

Answer:

the answer is C

Explanation:

This is because group6 elements are diatomic and when they are chemically combined their subscript 2 cancels out

6 0
2 years ago
A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solut
leonid [27]

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.

3 0
3 years ago
Which of the following is an example of a polymer?
svetlana [45]
<span>Sucrose </span>is an example of a polymer separate but equal means that both races get the same thing they just dont get it at the same time.<span />
3 0
3 years ago
How many elements had been discovered by 1860
UkoKoshka [18]

60 elements! I don't exactly know every element known, so if you are asking that, please tell us, but what I do know is that there were 60 elements in the periodic table by 1860.

5 0
3 years ago
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