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4 years ago

A box (with two ropes attached to it) rests on a table. Bob pulls to the left with

Chemistry

1 answer:

Anton [14]4 years ago

6 0

Answer:

50 N

Explanation:

it evens out at 200 N but because she's pulling with 250 her side pulls with 50 meaning it would be a pulling force of 50

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Argon occupies a volume of 28.3 L at a pressure of 0.36 atm. Find the volume of argon when the pressure is increased to 0.85 atm

FromTheMoon [43]

Answer:

Explanation:

Boyles law

P1 (.36)

V1(28.3)

P2(.85)

V2(?)

P1*V1=P2*V2

plug it in, and you get 12

3 0

3 years ago

How many protons does the biggest element of the periodic table have?

AnnZ [28]

Cesium is the biggest, atom wise.It has 55 protons.

3 0

3 years ago

What is the molality of a solution prepared by dissolving 75.8 g of ethylene glycol, HOCH2CH2OH, in 1.55 L of water? Assume the

GaryK [48]

Answer:

0.787 mol.Kg⁻¹ or 0.787 m

Explanation:

Data Given:

Mass = m = 75.8 g

Volume = V = 1.55 L = 1550 mL

M.Mass of Ethylene Glycol = 62.07 g/mol

Density of water = d = 1.0 g/mL

To Find:

Molality = ??

Molality is a unit of concentration and is defined as the number of moles of solute per Kg of a solvent.

Molality = Moles of Solute / Kg of Solvent --- (1)

First Calculate Moles of Solute as;

Moles = Mass / M.Mass

Moles = 75.8 g / 62.07 g/mol

Moles = 1.22 moles

Secondly calculate mass of water as,

mass of water = density × volume

mass of water = 1.0 g/mL × 1550 mL

mass of water = 1550 g or 1.55 Kg

Now, putting values of moles and mass of solvent in equation 1,

Molality = 1.22 mol / 1.55 Kg

Molality = 0.787 mol.Kg⁻¹ or 0.787 m

7 0

4 years ago

A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.

NeTakaya

Answer:

solubility of X in water at 17.0 $^{0}\textrm{C}$ is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 $^{0}\textrm{C}$ can be calculated using the information given.

Let's assume solubility of X in water at 17.0 $^{0}\textrm{C}$ is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

or, y = $\frac{3.96}{36}$ = 0.11

Hence solubility of X in water at 17.0 $^{0}\textrm{C}$ is 0.11 g/mL.

3 0

3 years ago

Acetic acid is a polar molecule and can form hydrogen bonds with water molecules. therefore, it has high solubility in water. ye

Keith_Richards [23]

Colligative properties calculations are used for this type of problem. Calculations are as follows:

Look up the freezing point of benzene, then delta T = f.p.=3.5 = Kf*m
Solve for molality. 

molality = moles/kg solvent 
Solve for moles. 

moles acetic acid = grams acetic acid/molar mass acetic acid 
Solve for molar mass. You would expect to find 60 for the molar mass CH3COOH.

4 0

3 years ago