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2 years ago

What is the connection between physical properties and physical changes

2 answers:

5 0

Answer: Physical changes are related to physical properties since some measurements require that changes be made. Melting Point: As solid matter is heated it eventually melts or changes into a liquid state at the melting point. Ice (a solid form of water) melts at 0 oC and changes to the liquid state.

Explanation:

Allisa [31]2 years ago

3 0

Answer:

Info from G00gle:Physical changes are related to physical properties since some measurements require that changes be made. Melting Point: As solid matter is heated it eventually melts or changes into a liquid state at the melting point. Ice (a solid form of water) melts at 0 oC and changes to the liquid state.

Explanation:

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A vector always consists of

bija089 [108]

Hope this helps A vector always consists of a direction and magnitude. F

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3 years ago

Nh4i(aq)+koh(aq)→ express your answer as a chemical equation. identify all of the phases in your answer.

tester [92]

NH4I (aq) + KOH (aq) in chemical equation gives

NH4I (aq) + KOH (aq) = KI (aq) + H2O(l) + NH3 (l)

Ki is in aqueous state H2o is in liquid state while NH3 is in liquid state

from the equation above 1 mole of NH4I (aq) react with 1 mole of KOH(aq) to form 1mole of KI(aq) , 1mole of H2O(l) and 1 Mole of NH3(l)

5 0

3 years ago

Read 2 more answers

What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 15.0% NaOH by mass?

GalinKa [24]

Hello!

a) The mole fraction of solute of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams of water. We can calculate the number of moles of each one in the following way:

$molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH \\ \\ moles H_2O=85 gH_2O* \frac{1 mol H_2O}{18 g H_2O}=4,7222 moles H_2O$

To finish, we calculate the mole fraction by dividing the moles of NaOH between the total moles:

$X_{NaOH}= \frac{moles NaOH}{total moles}= \frac{0,3750 moles NaOH}{0,3750 moles NaOH+4,7222 molesH_2O} =0,073$

So, the mole fraction of NaOH is 0,073

b) The molality (moles NaOH/ kg of solvent) of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams (0,085 kg) of water. We can calculate the moles of NaOH in the following way:

$molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH$

Now, we apply the definition of molality to calculate the molality of the solution:

$mNaOH= \frac{moles NaOH}{kg_{solvent}}= \frac{0,3750 moles NaOH}{0,085 kg H_2O}=4,41 m$

So, the molality of this solution is 4,41 m

Have a nice day!

4 0

3 years ago

Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units.

Thepotemich [5.8K]

Answer:

m = 2.955x10⁻² mol/kg

X = 5.323x10⁻⁴ mol NaCl/Total moles

(w/w)% = 0.1726%

ppm = 1726 mg/kg

Explanation:

Molality is the ratio between moles of solute per kg of solution.

As the solution is 2.950×10⁻² mol/L, mililters are 999.2mL and density is 0.9982 g/mL, molality is:

m = 2.950×10⁻² mol/L×(1L/0.9982kg) = 2.955x10⁻² mol/kg

Mole fraction is moles of NaCl/total moles.

Moles of H₂O are:

999.2mL×(0.9982g/mL)×(1mol/18,02g) = 55,35 moles of H₂O

Moles of NaCl are:

2.950×10⁻² mol/L×(0.9992L)= 2.950×10⁻² mol of NaCl

mole fraction is:

X = 2.950×10⁻² mol of NaCl / (2.950×10⁻² mol of NaCl+55.35mol water) = 5.323x10⁻⁴ mol NaCl/Total moles

Mass of NaCl is:

2.950×10⁻² mol of NaCl×(58.44g/mol) = 1.724g of NaCl

Mass of water is:

55.35mol water×(18.02g/mol) = 997.4g of H₂O

(w/w)% is:

1.724g of NaCl / (1.724g of NaCl+997.4g of H₂O) ×100 = 0,1726%

Parts per million is mg of NaCl per kg of solution, that is:

1724mg of NaCl / 0.999124g = 1726 ppm

I hope it helps!

5 0

3 years ago

A sample of argon (Ar) gas has a volume of 7.31 L at a pressure of

Pie

Answer:

17.1 L

Explanation:

P1V1/T1 = P2V2/T2 T is in K

P1V1/T1 * T2/P2 = V2

735 * 7.31 / (45+273.15) * (55+ 273.15)/325 = V2

V2 = 17.1 L

3 0

2 years ago