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3 years ago

14

What is the mass of a 1.00 l sample of a liquid that has a density of 0.967 g/ml ? express your answer using three significant f

igures?

1 answer:

BARSIC [14]3 years ago

5 0

Mass = 1000 mL x 0.967 g/1 mL = 967 g

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Who is the father of chemistry

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Answer: Antoine Lavoisier

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The volume of a gas filled balloon is 22L at 313K and 1.5 atm. What would the volume be at 273K and 1 atm?

avanturin [10]

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From the calculations and the statement of the general gas equation, the volume of the gas becomes 28.8 L

<h3>What are gas laws?</h3>

The gas laws are used to show the relationship between the variables in problems that has to do with gases.

In this case, we have to apply the general gas law as follows;

P1V1/T1 = P2V2/T2

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How would you be able to use the spectrum of each gas to establish its identity

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3 years ago

A 5-gal, cylindrical open container with a bottom area of 120 square inches is filled with glycerin and rests on the floor of an

aleksandrvk [35]

Answer:

a) 68.9 b/ ft²

b) 57.416

Explanation:

Volume of the cylindrical open container V = 5 gal

= (5)(231)

= 11155 in³

different diagrammatic expression has been shown in the attached file below for clear understanding.

Area A = 120 in²

V = hA

= $\frac{1155}{120}$

= 9.625 in

Upward acceleration , $a_z= 3ft/s^2$

$\frac{dp}{dz} = \rho (g + a_z)$

$dp= -\rho (g + a_z)dz$

$\int\limits^P}_0 \,dp= -\rho (g + a_z) \int\limits^0_h \, dz$

$P_b =$ $-\rho (g + a_z) h$

$= (2.4)*(32.2+3) (\frac{9.625}{12})$

= 68.9 b/ ft²

b) The flowchart for the resultant force is shown in the diagram below:

with that:

$F_f = P_bA$

$F_f =\frac{(68.9)(120)}{(144)}$

$F_f =57.416$

Thus, the resultant force that the container exert on the floor of the elevator during this acceleration = 57.416

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3 years ago

Consider a galvanic cell based on the reaction Al^3+_(aq) + Mg_(s) rightarrow Al_(s) + Mg^2+ _(aq) The half-reactions are Al^3+

grin007 [14]

<u>Answer:</u> The standard cell potential of the cell is -0.71 V

<u>Explanation:</u>

The half reactions follows:

<u>Oxidation half reaction:</u> $Mg\rightarrow Mg^{2+}+2e^-;E^o_{Mg^{2+}/Mg}=-2.37V$ ( × 3)

<u>Reduction half reaction:</u> $Al^{3+}(aq.)+3e^-\rightarrow Al(s);E^o_{Al^{3+}/Al}=-1.66V$ ( × 2)

The balanced cell reaction follows:

$2Al^{3+}(aq.)+3Mg(s)\rightarrow 2Al(s)+3Mg^{2+}(aq.)$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation, we get:

$E^o_{cell}=-2.37-(-1.66)=-0.71V$

Hence, the standard cell potential of the cell is -0.71 V

8 0

3 years ago