A group of students observed a sample of soil and recorded their observations.

Gasoline has a density of 0.749 g/ml. how many pounds does 19.2 gallons of gasoline weigh?

Svetradugi [14.3K]

A good first step is writing the amount in terms of ml.

19.2 gallons = 72.68 L = 72680 ml

that would mean it weighs 0.749*720680g = 54437.32ml = 54.437 L

hope that helps :)

6 0

2 years ago

Read 2 more answers

Suppose a liquid level from 5.5 to 8.6 m is linearly converted to pneumatic pressure from 3 to 15 psi. What pressure will result

wlad13 [49]

Answer:

a) P = 9.58 psi for h=7.2 m

b) P=4.7 psi for h=5.94 m

Explanation:

Since the pressure Pon a static liquid level h is

P= p₀ + ρ*g*h

where p₀= initial pressure , ρ=density , g = gravity

then he variation of the liquid level Δh will produce a variation of pressure of

ΔP= ρ*g*Δh → ΔP/Δh = ρ*g = ( 15 psi - 3 psi) /( 8.6 m - 5.5 m) = 12/3.1 psi/m

if the liquid level is converted linearly

P = P₁ + ΔP/Δh*(h -h₁)

therefore choosing P₁ = 3 psi and h₁= 5.5 m , for h=7.2 m

P = 3 psi + 12/3.1 psi/m *(7.2 m -5.5 m) = 9.58 psi

then P = 9.58 psi for h=7.2 m

for P=4.7 psi

4.7 psi = 3 psi + 12/3.1 psi/m *(h -5.5 m)

h = (4.7 psi - 3 psi)/ (12/3.1 psi/m) + 5.5 m = 5.94 m

then P=4.7 psi for h=5.94 m

5 0

3 years ago

A tank of oxygen has a volume of 1650 L. The temperature of the gas inside is 35?C. If there are 9750 moles of oxygen in the tan

Paul [167]

Answer:

2192.64 PSI.

Explanation:

From the general law of ideal gases:

PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the container in L (V = 1650 L).

n is the no. of moles of the gas in mol (n = 9750 mol).

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature of the gas in (T = 35°C + 273 = 308 K).

∴ P = nRT/V = (9750 mol)(0.082 L.atm/mol.K)(308 K)/(1650 L) = 149.2 atm.

To convert from atm to PSI:

1 atm = 14.696 PSI.

∴ P = 149.2 atm x (14.696 PSI/1.0 atm) = 2192.64 PSI.

4 0

2 years ago

in heating a kettle of water on an electric stove, 3.34×10^3 J of thermal energy was provided by the element of the stove. yet,

insens350 [35]

Answer:

The percentage efficiency of the electrical element is approximately 82.186%

Explanation:

The given parameters are;

The thermal energy provided by the stove element, $H_{supplied}$ = 3.34 × 10³ J

The amount thermal energy gained by the kettle, $H_{absorbed}$ = 5.95 × 10² J

The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;

$\eta \% = \dfrac{H_{supplied} - H_{absorbed} }{H_{supplied}} \times 100$

Therefore, we get;

$\eta \% = \dfrac{3.34 \times 10^3 - 5.95 \times 10^2}{3.34 \times 10^3} \times 100 = \dfrac{549}{668} \times 100 \approx 82.186 \%$

The percentage efficiency of the electrical element, η% ≈ 82.186%.

4 0

2 years ago

Which of the following is a strong base?

malfutka [58]

C. NaOH ammmonia is also an base but not as strong as NaOH

3 0

2 years ago