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yKpoI14uk [10]
2 years ago
8

a pharmacist has 10 and 90 iodine solutions on hand. how many liters of each iodine solution will be required to produce 8 liter

s of 50% iodine mixture
Mathematics
1 answer:
Westkost [7]2 years ago
6 0
1 litre of 10% solution contains 100mL of iodine
1 litre of 90% solution contains 900mL of iodine
Mixing them will produce 2 litres of solution containing 1000mL of iodine which is 50% iodine.
To produce 8 litres of 50% iodine they can mix 4 litres of 10% and 4 litres of 90%
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3.1 × 10^5

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Two-digit natural numbers are formed, with replacement, from the digits 0 through 9.
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THE AREA OF A LAKE ON THE MAP IS 12CM SQUARE CALCULATE THE ACTRUAL AREA OF THE LAKE IN KM
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if 1cm is for 1km there are 12km

Step-by-step explanation:

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6 0
3 years ago
The equation 7^2=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
jeyben [28]

Answer:

The period of Y increases by a factor of k^ {3/2} with respect to the period of X

Step-by-step explanation:

The equation T ^ 2 = a ^ 3 shows the relationship between the orbital period of a planet, T, and the average distance from the planet to the sun, A, in astronomical units, AU. If planet Y is k times the average distance from the sun as planet X, at what factor does the orbital period increase?  

For the planet Y:  

T_y ^ 2 = a_y ^ 3

For planet X:  

T_x ^ 2 = a_x ^ 3

To know the factor of aumeto we compared T_x with T_y

We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:  

a_y ^ 3 = (a_xk) ^ 3

So

\frac{T_y ^ 2}{T_x ^ 2}=\frac{a_y ^ 3}{a_x^ 3}\\\\\frac{T_y ^ 2}{T_x ^ 2}=\frac{(a_xk)^3}{a_x ^ 3}\\\\\frac{T_y^ 2}{T_x^ 2}=\frac{k ^ {3}a_{x}^ 3}{a_{x}^ 3}\\\\\frac{T_{y}^ 2}{T_{x}^ 2}=k ^ 3\\\\T_{y}^ 2 = T_{x}^{2}k^{3}\\\\T_{y} =k^{\frac{3}{2}}T_x

Then, the period of Y increases by a factor of k^ {3/2} with respect to the period of X



4 0
4 years ago
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