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sesenic [268]
3 years ago
7

The mass and volume of each sample differ from the mass and volume of the other samples. Is it possible for each sample to conta

in 1 mol of each substance?
A) No, because they have different masses.
B) No, because they have different volumes.
C) Yes, because the number of moles is not dependent on the mass or the volume.
D) Yes, because the number of moles is only dependent on the mass per unit volume.
Chemistry
2 answers:
sesenic [268]3 years ago
8 0

Answer: The correct answer is Option C.

Explanation:

To calculate the number of moles, we use the formula:

\text{number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given in the question that the masses of twp samples are different and from the above formula, we see that the moles of a substance depends on the molar masses of the substance.

Volume of the substance has no effect on the moles.

So, the two substances can have 1 mol of substance.

andrew11 [14]3 years ago
5 0
<span>The mass and volume of each sample differ from the mass and volume of the other samples. Is it possible for each sample to contain 1 mol of each substance?

</span><span>C) Yes, because the number of moles is not dependent on the mass or the volume.</span>
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A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a
deff fn [24]

Answer:

A) E_{a} = 350KJ/mol, E_{a} = 50KJ/mol, E_{a} = 50KJ/mol

     A = 1.5×10^{-7}s^{-1}, A = 1.9×10^{-7} s^{-1}, A=1.5×10^{-7} s^{-1}

B) 4.469

Explanation:

From Arrhenius equation

      K=Ae^{\frac{E_{a} }{RT} }

where; K = Rate of constant

            A = Pre exponetial factor

            E_{a} = Activation Energy

             R = Universal constant

             T = Temperature in Kelvin

Given parameters:

E_{a} =165KJ/mol

T_{1}=505K

T_{2}=525K

R=8.314JK^{-1}mol^{-1}

taking logarithm on both sides of the equation we have;

InK=InA-\frac{E_{a} }{RT}

since we have the rate of two different temperature the equation can be derived as:

In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )

In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1}  }.(\frac{1}{505} -\frac{1}{525} )

In(\frac{K_{2} }{K_{1} } )= 19846.04×7.544×10^{-5} = 1.497

\frac{K_{2} }{K_{1} } =e^{1.497} = 4.469

 

6 0
3 years ago
Can someone help me with number 1 and 2 plz!
GalinKa [24]

Answer:

1) 0 N

2) 8 N

Explanation:

The net force is the sum of all of the forces acting on the object.

For question 1, we can see that there is a force of 5 N acting to the right and 5 N acting to the left.  If we define the right to be positive and the left to be negative, then the net force equals:

Fnet = 5N - 5N = 0 N

Therefore, the net force in question 1 is 0 N.

For question 2, the process is very similar.  We want to find the sum of the forces acting on the object.  In this case, there are forces of 3 N and 5 N acting to the right.

Fnet = 3 N + 5 N = 8 N

Therefore, the net force in question 2 is 8 N.

Hope this helps!

3 0
3 years ago
A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.
MArishka [77]

Answer:

A. 85.6 g

= 0.0856 kg.

B. 0.00027 mol/g

= 0.27 mol/kg.

C. 8.39 %

Explanation:

Given:

Molar concentration = 0.25 M

Molar weight of sucrose = 342.296 g/mol

Density of solution = 1.02 g/mL

Mass of water = 934.4 g.

Density in g/l = 1.020 g/ml * 1000ml/1 l

= 1020 g/l

Mass of solution in 1 l of solution = 1020 g

Mass of solution = mass of solvent + mass of solute

Mass of sucrose = 1020 - 934.4

= 85.6 g of sucrose in 1 l of solution.

A.

Density of sucrose = mass/volume

= molar mass/molar concentration

= 342.296 * 0.25

= 85.6 g/l

Number of moles = mass/molar mass

= 85.6/342.296

= 0.25 mol

B.

Molality = number of moles of solute/mass of solvent

= 0.25/934.4

= 0.00027 mol/g

C.

% mass of sucrose = mass of sucrose/total mass of solution * 100

= 85.6/1020 * 100

= 8.39 %

6 0
3 years ago
Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a
Crank

Explanation:

a. Pb(NO_3)_2(aq) + Na_2SO_4(aq) → ?

Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)

Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)

Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)

Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)

b. NiCl_2(aq) + NH_4NO_3(aq) →

NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)

No precipitation is occuring.

c. Fe_Cl2(aq) + Na_2S(aq) →

FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)

FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)

Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)

Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)

d.MgSO_4(aq) + BaCl_2(aq) →

MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2

MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)

BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)

Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

5 0
3 years ago
It is recommended that drinking water contain 1.6 ppm fluo- ride (F) to prevent tooth decay. Consider a cylindrical reservoir wi
Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d =4.50\times 10^1 m=45 m

r = d/2 = 22.5 m

Depth of the reservoir = h =  10.0 m

V=\pi r^2 h

=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L

1 m^3=1000 l

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir =  m

m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then 15,896,250 kg of water have x mass of fluorine:

\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}

x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434  grams of fluorine  should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%

Total mass of hydrogen hexafluorosilicate = m'

79.16\%=\frac{25,434 g}{m'}\times 100

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

5 0
3 years ago
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