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3 years ago

How many milliliters of a 3.0 M HCL solution are required to make 250.0 millimeters of 1.2 M HCL?

1 answer:

7 0

The problem above can be solved using M1V1=M2V2 where M1 is the concentration of the concentrated, V1 is the volume of the concentrated solution, M2 is the concentration of the Dilute Solution, V2 is the Volume of the dilute solution. Hence,

(3.0 M)(V2)=(250 mL)(1.2M)
V2 (3.0)= 300
V2= 100 mL

Therefore, you need 100 mL of 3.0 M HCl to form a 250 mL of 1.2 M HCl.

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3 years ago

A and b are two gases that are mixed together: 2.50 mol a is mixed with 0.850 mol b. if the final pressure of the mixture is 1.7

USPshnik [31]

Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.

The partial pressure of a gas in a mixture can be calculated as

Pi = Xi x P

Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.

Therefore we have Pa = Xa x P and Pb = Xb x P

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6 0

2 years ago

True or false

Kaylis [27]

Answer:

True

Explanation:

Forces, equal in magnitude and opposite in direction, and cancel each other out are called Balanced Forces.

Balanced Forces DO NOT cause any change in the velocity of the object, but the Object may correspond to a change in shape or size.

Example :

Blasting a balloon by compressing it tightly with both hands, roughly with an equal force, is a practical example of Balanced Forces. In this case, the balloon before the burst stays at a state of rest and same after the burst. But the burst or the compression, causes a change in the shape of the balloon.

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3 years ago

Calculate the molality of a solution prepared by dissolving 175 g of KNO3 in 750 g of water. (round to nearest hundreth)

Wittaler [7]

We will get the molality from this formula:
Molality = no.of moles of solute / Kg of solvent

So first we need the no.of moles of KNO3 = the mass of KNO3 / molar mass of KNO3
no.of moles of KNO3 = 175 / 101.01 = 1.73 mol
By substitution in the molality formula:
∴ molality = 1.73 / (750/1000) = 2.3 Molal

6 0

3 years ago

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