Now liters at the same temperature and pressure: liters of H2 + liters of O2 → liters of H2O

A fluid occupying has a mass of 4mg. Calculate its density and specific volume in SI, EE, and BG units.

kondaur [170]

The question is incomplete, complete question is:

A fluid occupying $3.2 m^3$ of volume has a mass of 4 Mg. Calculate its density and specific volume in SI, EE, and BG units.

Explanation:

1) Mass of liquid = m = 4 Mg = 4 × 1,000 kg = 4,000 kg

(1 Mg = 1000 kg)

Volume of the fluid = V = $3.2 m^3$

Density of the fluid = D

$D=\frac{m}{V}=\frac{4,000 kg}{3.2 m^3}=1,250 kg/m^3$

Specific volume is the reciprocal of the density :

$V_{specific}=\frac{1}{Density}$

Specific volume of the fluid = $S_v$

$S_v=\frac{1}{D}=\frac{1}{1,250 kg/m^3}=0.0008 m^3/kg$

2)

Density of the fluid in English Engineering units = D $(lb/ft^3)$

1 kg = 2.20462 lb

1 m = 3.280 ft

$D=\frac[1,250\times 2.20462 lb}{(3.280 ft)^3=78.95 lb/ft^3$

Specific volume of the fluid :

$=\frac{1}{78.95 lb/ft^3}=0.0127 ft^3/lb$

3)

Density of the fluid in British Gravitational System units = D $(slug/ft^3)$

1 kg = 0.06852 slug

1 m = 3.280 ft

$D=\frac[1,250\times 0.0685218 slug}{(3.280 ft)^3=2.43 slug/ft^3$

Specific volume of the fluid :

$=\frac{1}{2.43 slug/ft^3}=0.412 ft^3/slug$

7 0

3 years ago

Now, examine the structures of benzhydrol and fluorene. Both compounds contain the same number of carbons but have very differen

zubka84 [21]

Answer:

The OH group

Explanation:

Benzhydrol contains OH hydroxyl group in its molecule while fluorene does not. At first glance, one would think that OH, which contributes to hydrogen bonding would causes melting point of benzhydrol to be higher than fluorene. However, the structure of benzhydrol, which is 2 benzene rings connected to center hydroxyl carbon (PhCOHPh), allows for each benzene rings in benzhydrol to rotate until both rings are perpendicular to minimize repulsive force. This prevents the molecule from stacking on each other due to its non flat shape, and thus, lowering its melting point in contrast to flat fluorene molecule.

4 0

3 years ago

How do you make liquid nitrogen icecream

Harrizon [31]

Add any other liquid flavorings you might want. Put on your gloves and goggles. Pour a small amount of liquid nitrogen directly into the bowl with the ice cream ingredients. Continue to stir the ice cream, while slowly adding more liquid nitrogen.

7 0

3 years ago

Read 2 more answers

What will happen to the chemical equilibrium if NH4Cl is added to this solution?

Tresset [83]

The answer to this item depends entirely to the chemical reaction. If the compound, NH4Cl, is in the left hand side of the reaction, when it is added, the reaction will shift to the left. In the same manner, when the compound is in the right-hand side of the reaction, the reaction will shift to the right.

This happens because initially the reaction is in equilibrium and adding another compound to it will most likely lead to the shifting of the reaction.

6 0

3 years ago

Read 2 more answers

The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5

Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

4 0

3 years ago