CHEMISTRY MIDDLE SCHOOL

Solutes lower the freezing point of water by: a. Stopping water
molecules from moving

b.forming crystals

c.making it harder for water to form crystals

d. Making the water molecules move faster

Answers

Answer 1
Answer: Water molecules are constantly moving in Ice they move slower and are more tightly packed together and move at a slower pace to lower the freezing point you would have to find a solute that makes it harder for water to form crystals 

Related Questions

HIGH SCHOOL

when curium 242 is bombarded with an alpha particle two products are formed one of which is a neutron what is the other product

Answers

Lol your in my science class. i believe its plutonium-239, but if someone else says I'm wrong listen to them. I'm not too sure.

It is actually Californium-245 that is formed.

COLLEGE

Calculate the molar mass of aluminum oxide (Al2O3). Express your answer to four significant figures. The molar mass of Al2O3 is
grams.

Answers

Answer:

102.0 g/mol

Explanation:

HIGH SCHOOL

How can molecules with polar bonds be nonpolar?

Answers

A molecule can possess polar bonds and still be non polar. If the polar bonds are evenly (or symmetrically) distributed, the bond dipoles cancel and do not create a molecular dipole.
MIDDLE SCHOOL

WILL GIVE BRAINLIEST!!! NEED ANSWER RIGHT NOW!!!!! Which of the following statements is true?

On a level surface, applied force equals weight.
A Newton is equal to 9.8 m/s^2.
Normal forces occur at perpendicular angles to the surface between two objects.
Action-at-a-distance forces require physical contact between objects.

Answers

Normal forces occur at perpendicular angles to the surface between two objects.

Option B

Explanation:

The unit of force comes from its mathematical equation which says force is equal to the product of mass and acceleration. So, the unit of force becomes kg m/s². And this is the SI unit of force, which is also called newton. So 1 newton = 1 kgm/s².

Action at a distance forces are actually the non contact forces which can even work on a body even when the bodies aren't in contact with each other. The examples include gravitational force, magnetic and electric forces etc. So they don't need a physical contact between the two bodies.

But, normal forces are defined as the force that's acting directly on the axis of two bodies i.e perpendicularly to a surface or the part of a surface the body is resting. So, they have full impact on the touching surface.

Answer:

Normal forces occur at perpendicular angles to the surface between two objects.

HIGH SCHOOL

You now know that [Pb+2] = 0.17 M. Calculate [Cl-]. Refer to equations given in the text to determine the relationship between these ions.

0.17 M
0.26 M
0.34 M
0.51 M

Answers

Answer:

0.34 M

Explanation:

Concentration of Pb2+ = 0.17 M

Pb2+ (lead ion) and Cl- (chloride ion) will combine to form lead chloride (PbCl₂)

Based on the chemical formula stoichiometry for every Pb2+ there are 2 Cl-.

Therefore, the relationship between the two ions would be:

[Cl-] = 2([Pb2+])

[Cl-] = 2(0.17 M) = 0.34 M

The correct answer would be .34 M
HIGH SCHOOL

What is the smallest unit retaining properties of element?

Answers

Atom is the smallest unit retaining properties of elements
MIDDLE SCHOOL

Whats a covalent polar chemical bond

Answers

Polar covalent bonding is a type of chemical bond where a pair of electrons is unequally shared between two atoms. In a polar covalent bond, the electrons are not equally shared because one atom spends more time with the electrons than the other atom.
:)
COLLEGE

Use Henry's law and the solubilities given below to calculate the total volume of nitrogen and oxygen gas that should bubble out of 1.0 L of water upon warming from 25 ∘C to 50 ∘C. Assume that the water is initially saturated with nitrogen and oxygen gas at 25 ∘C and a total pressure of 1.0 atm. Assume that the gas bubbles out at a temperature of 50 ∘C. The solubility of oxygen gas at 50 ∘C is 27.8 mg/L at an oxygen pressure of 1.00 atm. The solubility of nitrogen gas at 50 ∘C is 14.6 mg/L at a nitrogen pressure of 1.00 atm. Assume that the air above the water contains an oxygen partial pressure of 0.21 atm and a nitrogen partial pressure of 0.78 atm.

Answers

Answer:

Explanation:

Henry's law relates the solubility of a gas in a solvent to it's partial pressure above the solution. P=k*solubility, where P is the pressure and k is the Henry's law constant. The Henry's law constants are temperature dependent.

There's basically three steps here (I'll just talk about oxygen here, the procedure for nitrogen is analogous). You first need to calculate how much oxygen is dissolved at 25 C. This is what you need the room temperature Henry's law constant for. You can then plug the constant and partial pressure (.21 atm) into Henry's law to get the concentration, and that can be converted into amount since you know the volume of water (1.0 L).

Alright, next you need to calculate how much oxygen will be dissolved at 50 C. You can do this by first finding the Henry's law constant, which you can do since you know the solubility at 1.00 atm and can plug that into Henry's law (k*27.8 mg/L=1.00 atm), and then use that to figure out the concentration at a pressure of .21 atm. And then translate that to amount of oxygen.

So now you know how much oxygen is dissolved at 25 C, and how much oxygen will be dissolved in 50 C. So, obviously, the difference is how much oxygen is released; translate this into volume using the ideal gas law to figure out what the volume of that amount of oxygen is.

Be careful with units throughout, that may well be the trickiest part.

Answer:

Explanation:

Given:

  • At 25∘C and total pressure of 1.0 atm,
  • Partial pressure of Oxygen, PO2 = 0.21 atm and,
  • Partial pressure of nitrogen, PN2 = 0.78 atm.

At 50∘C and 1 atm,

  • Solubility of oxygen = 27.8 mg/ L and solubility of nitrogen = 14.6 mg/ L

Apply the Henry law, Pi = H x Ci.................equation1

  • Where Pi is the partial pressure of species 'i' and H is Henry constant and Ci is the solubility of species 'i'., Henry constant for oxygen is HO2 = 769.23 L atm/mol

From the equation (1) the solubility of oxygen at 25∘C;

  • = 0.21/769.23 = 0.000273mol/L
  • or C'(O2) = 0.000273 X 32g X gL/mol X mol/L = 8.736mg

Therefore ; the solubility of oxygen at 50oC and pressure @ 1 atm,

  • = 27.8 X 1L X mg/L = 27.8mg

  • but, the partial pressure of oxygen in the solution is 0.21 atm, hence the solubility will be decreased to 27.8 x 0.21 atm/1 atm = 5.838mg
  • from 25∘C - 50∘C ; 8.736 - 5.838 = 2.898 mg of oxygen should be bubble out.

Hence, Volume of oxygen = Mass/density = 2.898mg/1.429mg/L = 2.027mL

  • Similarly, For nitrogen, At 50∘C and 1 atm,
  • solubility of nitrogen = 14.6 mg/ L
  • Hence, Henry constant for Nitrogen is HN2 = 1639.34 L atm/mol,

From the equation (1) the solubility of nitrogen at 25oC ;

  • C(N2) = 0.78/1639.34 = 0.0004758mol/L
  • or Ci = 0.0004758mol/L X 28g X 1L/mol X mol/L = 13.322mg

Hence, The solubility of Nitrogen at 50oC and pressure @ 1 atm,;

  • Ci = 14.6 X 1L X mg/L = 14.6mg

But the partial pressure of Nitrogen in the solution is 0.78 atm, hence the solubility will be reduced to 14.6 x 0.78 atm/1 atm = 11.388 mg

warming from 25∘C - 50∘C; 13.322 - 11.388 = 1.934 mg of Nitrogen

  • Hence, volume of nitrogen = mass/density = 1.934mg/1.2506mg/L
  • = 1.55mL

MIDDLE SCHOOL

the directions in the candy recipe for pralines instruct the cook to remove the pot containing the candy mixture from the heat when the candy micture reaches the soft-ball stage. the soft ball stage corresponds to the temperature of 236°F. after the soft ball stage is reached ,the pecans and vanilla are added. can a celsius thermometer with a range of -10°C be used to determine when the soft ball stage is reached in the candy micture? why or why not?

Answers

A Celsius thermometer with a range of -10°C cannot be used to determine when the soft ball stage is reached in the candy mixture.  

Temperature is most commonly measured with a thermometer that has two different scales: Fahrenheit and Celsius. Water freezes at 0°C and boils at 100°C on the Celsius scale and freezes at 32°F and boils at 212°F on the Fahrenheit scale.

Temperature of 236°F corresponds to 113.3°C. Now a Celsius scale has a maximum of 100°C on the thermometer. Hence a Celsius thermometer cannot be used to measure a temperature that corresponds to 236°F.  


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