What was Ernest Rutherford's experiment?

A certain block is floating in water with 77% of itself submerged. It has a length of 4.3 cm, a width of 5.9 cm, and a height of

dmitriy555 [2]

Answer:

f. 0.0537 kg

Explanation:

To solve this, we use the formula for force of buoyancy (Fb) = Vs × D × g

Where; Fb= Force of buoyancy

Vs= Volume of submerged

block

D= Density of water that block

submerges in

g= acceleration due to gravity

In this question, volume of block can be calculated using L×B×H

= 4.3cm × 5.9cm × 2.75cm = 69.7675cm^3

If 77% of the block is submerged in water, then 77/100 × 69.7675cm^3

= 53.72cm^3 is the Volume of submerged block

Density of water is known as 1000kg/m^3 while the gravitational force is 10m/s^2

We need to convert the Vs (volume of submerged block) to m^3 by dividing by 1,000,000

Hence, Vs = 53.72/1000000

Vs= 0.00005372m^3

Using the formula; (Fb) = Vs × D × g

Fb = 0.00005372m^3 × 1000kg/m^3 × 10m/s^2

Fb = 0.5372kg/m/s^2

Therefore, the force of buoyancy of the object on water is 0.5372kg/m/s^2

Recall that; Mass (m) = Volume (V) × Density (D)

Force of buoyancy (Fb) = Vs×D = mass (m) × acceleration due to gravity (g)

i.e. mass = Fb/g

= Mass of the block = 0.5372/10

Mass = 0.05372kg

Therefore, the mass of the block in kg is 0.05372kg

7 0

2 years ago

What is the difference between the Lewis model and the valence-shell electron pair repulsion (VSEPR) model?

grandymaker [24]

Answer: only the VSEPR mode shows the geometric shape of a formula.

8 0

2 years ago

1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?

love history [14]

Answer:

$m=1.325gNa_2CO_3$

Explanation:

Hello,

In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:

$eq-g=0.5eq-g/L*0.050L=0.025eq-g$

Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:

$mol=0.025eq-gNa_2CO_3*\frac{1molNa_2CO_3}{2eq-gNa_2CO_3}\\ \\mol=0.0125molNa_2CO_3$

Finally, the mass is computed by using its molar mass (106 g/mol)

$m=0.0125molNa_2CO_3*\frac{106gNa_2CO_3}{1molNa_2CO_3} \\\\m=1.325gNa_2CO_3$

Regards.

7 0

2 years ago

An aqueous solution containing 17.5 g of an unknown molecular (nonelectrolyte) compound in 100.0 g of water has a freezing point

Sonbull [250]

Answer:

molar mass = 180.833 g/mol

Explanation:

mass sln = mass solute + mass solvent

∴ solute: unknown molecular (nonelectrolyte)

∴ solvent: water

∴ mass solute = 17.5 g

∴ mass solvent = 100.0 g = 0.1 Kg

⇒ mass sln = 117.5 g

freezing point:

ΔTc = - Kc×m

∴ ΔTc = -1.8 °C

∴ Kc H2O = 1.86 °C.Kg/mol

∴ m: molality (mol solute/Kg solvent)

⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)

⇒ m = 0.9677 mol solute/Kg solvent

molar mass (Mw) [=] g/mol

∴ mol solute = ( m )×(Kg solvent)

⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )

⇒ mol solute = 0.09677 mol

⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )

⇒ Mw solute = 180.833 g/mol

6 0

2 years ago

At 333 k, which of the pairs of gases below would have the most nearly identical rates of effusion?

rjkz [21]

E. co and n2Effusion is the process where gas escapes through a hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. Relative rates of effusion is related to the molecular mass.
a) M(N₂)/M(O₂) = 28/32 = 0,875
b) M(N₂O)/M(NO₂) = 44/46 = 0,956
c) M(CO)/M(CO₂) = 28/44 = 0,636
d) M(NO₂)/M(N₂O₂) = 44/58= 0,758
e) M(CO)/M(N₂) = 28/28 = 1, CO and N₂ have iexact molecular masses and will effuse at nearly identical rates.

8 0

2 years ago

What was Ernest Rutherford's experiment?​

What was Ernest Rutherford's experiment?