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mojhsa [17]
3 years ago
15

A student is told to use 20.0 g of sodium chloride to make an aqueous solution that has a concentration of 10.0 g/L (grams of so

dium chloride per liter of solution). Assuming that 20.0 g of sodium chloride has a volume of 7.50 mL, show that she will need about 1.99 L of water to make this solution. In making this solution, should she add the solute to the solvent or the solvent to the solute?
Physics
2 answers:
Stells [14]3 years ago
6 0

Answer:

she should add solute to the solvent

Explanation:

Given data :

Mass of the sodium chloride, = 20.0 g

Concentration of the solution = 10 g/L

Volume of 20.0 g of sodium chloride = 7.50 mL

Now, from the concentration, we can conclude that for 10 g of sodium chloride volume of the solution is 1 L

thus, for 20 g of sodium chloride  volume of the solution is 2 L or 2000 mL

also,

Volume of solution = Volume of solute(sodium chloride) + volume of solvent (water)

thus,

2000 mL = 7.5 mL + volume of solvent (water)

or

volume of water = (2000 - 7.5) mL

or

volume of water = 1992.5 mL

or

volume of water = 199.25 L ≈ 199 L

3241004551 [841]3 years ago
5 0

Explanation:

Concentration of sodium chloride = 10.0 g/L

1 Liters of solution has 10 g of sodium chloride

The volume of solution in which 20 g of sodium chloride will present is:

\frac{1 L}{10 }\times 20 =2 L

Volume of solution we will be making is of 2 L.

Volume of 20 g of sodium chloride = 7.50 ml = 0.00750 L

Volume of the water = ?

Volume of solution = Volume of solute + Volume of solvent

Volume of solvent = 2L - 0.00750 L = 1.9925 L

So,volume of water required to prepare the 2 L of 10 g/L solution of sodium chloride is 1.9925 L.

And for while making of the solution , student must add solute the solvent.

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Answer:

D. Exothermic, because energy is being absorbed from the surroundings

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This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

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a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

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Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

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Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

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the tangential acceleration of the ball will be;

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we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

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3 years ago
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Answer:

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Mathematically, it is given by the formula;

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Explanation:

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Answer:

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Thank you for reading.

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3 years ago
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