What determines the number of possible sublevels?

If kerosene has a specific gravity of 0.820, what force will be exerted on the circular bottom of a cylindrical kerosene tank th

Svet_ta [14]

Answer:

$F = 774146.534\,N$

Explanation:

The pressure at the bottom of the tank is:

$P_{bottom} = (0.820)\cdot (1000\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot (30\,ft)\cdot (\frac{0.305\,m}{1\,ft} )$

$P_{bottom} = 73581.921\,Pa$

The force exerted on the circular bottom is:

$F=(73581.921\,Pa)\cdot (\frac{\pi}{4} )\cdot [(12\,ft)\cdot (\frac{0.305\,m}{1\,ft} )]^{2}$

$F = 774146.534\,N$

4 0

3 years ago

PLEASE HELP REALLY IMPORTANT

USPshnik [31]

The temperature of the water in both beakers is greater than 8 °C
This is because since the water passes on heat to the spheres, that means they are at a higher temperature then the spheres. if they were at a lower temp. The spheres would pass on heat to the water.

3 0

3 years ago

What mass of iron is formed when 10 grams of carbon react with 80 grams of iron iii oxide?

yKpoI14uk [10]

Answer:

55.85 grams of Fe is formed.

Explanation:

Identify the reaction:

2Fe₂O₃ + 3C → 4Fe + 3CO₂

Identify the limiting reactant, previously determine the mol of each reactant

(mass / molar mass)

10 g / 12 g/m = 0.83 moles C

80 g / 159.7 g /m = 0.500 moles Fe₂O₃

2 moles of oxide need 3 moles of C, to react

0.5 moles of oxide, will need ( 0.5 . 3)/ 2 = 0.751 mol

I have 0.83 moles of C, so C is the excess.

The limiting is the oxide.

3 mol of C need 2 mol of oxide to react

0.83 mol of C, will need (0.83 . 2)/ 3 = 0.553 mol of oxide, and I only have 0.5 (That's why Fe₂O₃ is the limiting)

Ratio is 2:4 (double)

If I have 0.5 moles of oxide, I will produce the double, in the reaction.

1 mol of Fe, will be produce so its mass is 55.85 g

5 0

3 years ago

How many atoms are in 25.0 grams of carbon?

mars1129 [50]

mol = 25/12

= 2.083 mol

1 mol = 6.02 × 10^23 atoms

2.083 mol = X

X = 2.083/1 × 6.02 × 10^23

= 1.254 × 10^24 atoms

5 0

3 years ago

A 1.30M solution of BaCl2 has a density of 1.230 g/ml. a) What is the mole fraction of BaCl2 in this solution?

Elodia [21]

Answer: The mole fraction of barium chloride in the solution is 0.024

Explanation:

We are given:

Molarity of barium chloride solution = 1.30 M

This means that 1.30 moles of barium chloride is present in 1 L or 1000 mL of solution.

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.230 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.230g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.230g/mL\times 1000mL)=1230g$

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Moles of barium chloride = 1.30 moles

Molar mass of barium chloride = 208 g/mol

Putting values in equation 1, we get:

$1.30mol=\frac{\text{Mass of barium chloride}}{208g/mol}\\\\\text{Mass of barium chloride}=(1.30mol\times 208g/mol)=270.4g$

Mass of water = Mass of solution - Mass of barium chloride

Mass of water = 1230 - 270.4 = 959.6 g

Calculating the moles of water:

Given mass of water = 959.6 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{959.6g}{18g/mol}\\\\\text{Moles of water}=53.31mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

For barium chloride:

$\chi_{\text{(barium chloride)}}=\frac{n_{\text{(barium chloride)}}}{n_{\text{(water)}}+n_{\text{(barium chloride)}}}$

$\chi_{\text{(barium chloride)}}=\frac{1.30}{1.30+53.31}\\\\\chi_{\text{(barium chloride)}}=0.024$

Hence, the mole fraction of barium chloride in the solution is 0.024

6 0

4 years ago