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irakobra [83]
2 years ago
10

How much pure alcohol must a pharmacist add to 10 cm.

SAT
1 answer:
Lera25 [3.4K]2 years ago
3 0

The amount of pure alcohol that the pharmacist must add to 10 cm of a 10% alcohol solution to strengthen it to a 70% solution is; 20 cm³

<h3>Volume of Solutions</h3>

Let x represent the volume of 100% pure alcohol in cm³ that is required to be added. Thus;

x(100%) + 10%(10 cm³) = 70%(x + 10 cm³)

Multiplying out the bracket gives;

x + (10 × 0.1) = 0.7(x + 10)

⇒ x + 1 = 0.7x + 7

x - 0.7x = 7 - 1

0.3x = 6

x = 6/0.3

x = 20 cm³

In conclusion. the volume of pure alcohol required to be added to 10 cm of a 10% alcohol solution to strengthen it to a 70% solution is 20 cm³

The complete question is;

How much pure alcohol must a pharmacist add to 10 cm of a 10% alcohol solution to strengthen it to a 70% solution? ASAP

Read more about Volume of solution at; brainly.com/question/26015585

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<h3>What is an integral?</h3>

An Integral is a variable of which a given function is the derivative, i.e. it gives that function when differentiated and may express the area under the curve of the function's graph.

<h3>What is the explanation to above answer?</h3>

Given:

F(x) = 37 cox (x²)

Internal = [0,1] n = 8 in Δ x = 1/8

The sub intervals are:

[0, 1/8], [1/8, 2/8], [2/8, 3/8], [ 3/8, 4/8], [ 4/8, 5/8], [ 5/8, 6/8], [6/8, 7/8], [ 7/8, 1]

The mid points are given as:

1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16

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Using the Trapezium Rule which states that:

\int\limits^1_0 cos(x)^{2} } \, dx = Δx/2 [f(xo) + 2f(x1) 2f(x2) + ....+ 2f(x7) + f(x8)]

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T8 = 33.386321

It is to be noted that the midpoints rule is given as;

\int\limits^1_0 {Cos(x)^{2} } \, dx  = Δx [f(1/16) + (3/16) + .... + f(15/16)]

= 1/8[f(1/16) + f (3/16) + f(5/16) + f(7/16) + f(9/16) + f(11/16) + f(13/16) + f(15/16)]

= 0.905620

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