All categories

3 years ago

Something with a density greater than 1.00 g ml

1 answer:

5 0

Something with a density greater than 1.00 g ml will sink in waterThe density of water is 1 gm/cm^3=1000 kg/m^3, therefore, any element or object with density greater than 1 will sink in water.
Examples of elements with density greater than 1:
Aluminum (2700 kg/m^3 = 2.7 gm/cm^3)
Nickel ( 8910 kg/m^3 = 8.91 gm/cm^3)

You might be interested in

A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and

lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

$F_{solution}$ = mass of solvent + mass of oil (i.e A+C)

Feed Phase:

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

= 25kg

Mass ratio of oil to solution $Y_{F}$ = $\frac{Mass C}{Mass (A+C)}$

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg

therefore $Y_{F}$ = $\frac{2.5}{25}$

= 0.1

mass ratio of solid to solution $Y_{A}$ = $\frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}$

Solvent Phase:

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

Underflow:

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = $\frac{mass B}{mass(A+C)}$

solution in underflow E₁ = Mass (A+C)

Overflow:

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75= E₁ × N₁

75 = E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

5 0

3 years ago

Which of the following would happen if Earth's moon were half the size that it is now? O The oceans' tidal variations would be s

Brilliant_brown [7]

Answer: A. The oceans‘ tidal would be smaller because the moon would exert less gravitational pull on earths oceans.

Explanation:

i got it right :)

3 0

3 years ago

ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG o , and predict the direction in which the rea

kondaur [170]

Answer:

The reaction is not spontaneous in the forward direction, but in the reverse direction.

Explanation:

Step 1: Data given

H2(g) + I2(g) ⇌ 2HI(g) ΔG° = 2.60 kJ/mol

Temperature = 25°C = 25+273 = 298 Kelvin

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Step 2: Calculate ΔG

ΔG = ΔG° + RTln Q

with ΔG° = 2.60 kJ/mol

with R = 8.3145 J/K*mol

with T = 298 Kelvin

Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]

with pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Q = (3.10²)/(1.5*1.75)

Q = 3.661

ΔG = ΔG° + RTln Q

ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)

ΔG =5815.43 J/mol = 5.815 kJ/mol

To be spontaneous, ΔG should be <0.

ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.

4 0

3 years ago

What determines the type of crystals in igneous rocks?

nika2105 [10]

Over all the size of the crystal depends on how fast magma will harden over time if the magma cools slowly the crystal will be larger than they would if the magma cooled faster

7 0

3 years ago

Create and balance a chemical equation for the following statement.

m_a_m_a [10]

FeCl₂ + 2NH₄OH → Fe(OH)₂ + 2NH₄Cl

8 0

3 years ago