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FromTheMoon [43]

2 years ago

15

a graduated cylinder is filled to 41.5 mL with water and a piece of granite is placed in the cylinder, displacing the level to 4

7.6mL. What is the volume of the granite piece in cubic centimeters.

1 answer:

Maksim231197 [3]2 years ago

8 0

Answer:

The answer is

<h2>6.1 cm³</h2>

Explanation:

To find the volume in cm³ we must first find the volume of the rock .

To find the volume of the rock , subtract the initial volume of water from the final volume of the water when the rock was dropped in it

That's

volume = final volume of water - initial volume of water

volume of rock = 47.6 mL - 41.5 mL

= 6.1 mL

Next we use the conversion

<h3>1 mL = 1 cm³</h3>

If 1 mL = 1 cm³

Then 6.1 mL = 6.1 cm³

Hope this helps you

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What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the de

Snezhnost [94]

Here is the full question

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 10 cm. The two liquids do not mix.

What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.

Express your answer in two significant figures and include the appropriate units (in cm)

Answer:

ΔH ≅ 3.73 cm

Explanation:

The pressure inside a liquid is known as hydrostatic pressure and which is represent by the formula:

P = ρ × g × h

where;

ρ is the density of the fluid

g is the gravitational constant

h is the height from the surface

From the question above;

For glycerine; we have:

density of glycerine = 1260 kg/m³

gravitational constant = 9.8 m/s²

height = ???

∴

$P_{(g)= 1260kg/m^3}*9.8m/s^2*h_g$ ----- equation (1)

On the other hand for alcohol:

density of alcohol is given as = 790 kg/m³

gravitational constant = 9.8 m/s²

height = 10 cm

∴

$P_{(a)= 790kg/m^3*9.8m/s^2*10$ ----------- equation (2)

if we equate equation 1 and 2 together; we have

$P_{(g)= P_{(a)$

$1260kg/m^3}*9.8m/s^2*h_g = 790kg/m^3*9.8m/s^2*10cm$

Making $h_g$ the subject of the formula, we have :

$h_g= \frac{ 790kg/m^3*9.8m/s^2*10cm}{1260kg/m^3*9.8m/s^2}$

$h_g$ = 6.269 cm

The difference in the height denoted by ΔH can therefore be calculated as:

ΔH $= H_a-H_g$

ΔH $= 10cm - 6.269cm$

ΔH = 3.731 cm

ΔH ≅ 3.73 cm (to two significant figures)

5 0

2 years ago

Air tends to move from _____.

beks73 [17]

A: The Equator To The Poles

4 0

2 years ago

Read 2 more answers

A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Lit

My name is Ann [436]

Answer:

$M=0.213M$

Explanation:

Hello,

In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:

$n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-$

$n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-$

$n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-$

We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:

$n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-$

Finally, we compute the molarity:

$M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M$

Regards.

6 0

2 years ago

a drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25C. if the heat durin

Gwar [14]

Mass of water vapor is 0.48 gms.

Weight of the aluminium block is 55 gms.

Heat of vaporization of water at 25 degree Celsius is 44.0 kJ/mol.

The amount of heat given by the condensation is:

q = Heat of vaporization × Mass of vapor / Molar mass of steam

= 44.0 kJ/mol ×0.48 g / 18 g/mol

= 1.173 kJ

Now the final temperature of the metal block is calculated by the formula:

q = m × c × ΔT

q = m × c × (T₂ - T₁)

Here, q is the amount of heat, m is the mass of the metal, c is the specif heat of the metal, T₁ is initial temperature, and T₂ is the final temperature.

Now, substituting the values we get,

1.173 kJ = 55 g × 0.903 J/g. ° C * (T₂ - 25°C)

1.173 × 10³ J = 55 g × 0.903 J/g. degree C × (T₂ - 25°C)

1.173 × 10³ = 49.665 ° C * (T₂ - 25° C)

T₂ = 49° C.

Thus, the final temperature of the metal block is 49° C.

8 0

2 years ago

A certain gas is present in a 12.0 L cylinder at 4.0 atm pressure. If the pressure is increased to 8.0 atm the volume of the gas

diamong [38]

There are 3 parts in this question:
1) To find the initial Boyle's constant $k_{i}$
2) To find the final Boyle's constant $k_{f}$
3) To verify whether gas is obeying Boyle's law or not

Given data:
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm

The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L

First you need to know what Boyle's law is:
<span>Boyle's law states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.
</span>
The Mathematical form of Boyle's law is:
$P = \frac{k}{V}$

Where,
P = Pressure
V = Volume of the gas
k = Boyle's constant

Now let's solve aforementioned parts one by one:
1.
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm
The Boyle's constant = $k_{i}$ = ?

According to the Boyle's law,

$P_{i} = \frac{k_{i}}{V_{i}}$

=> $k_{i}$ = $P_{i}V_{i}$
Plug-in the values in the above equation, you would get:
$k_{i}$ = 4.0 * 12.0 = 48

Ans-1) $k_{i}$ = 48

2.
The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L
The Boyle's constant = $k_{f}$ = ?

According to the Boyle's law,

$P_{f} = \frac{k_{f}}{V_{f}}$

=> $k_{f}$ = $P_{f}V_{f}$
Plug-in the values in the above equation, you would get:
$k_{f}$ = 8.0 * 6.0 = 48

Ans-2) $k_{f}$ = 48

3.
In order to verify Boyle's law, the initial Boyle's constant should be EQUAL to the final Boyle's constant, meaning:

$k_{i}$ = $k_{f}$

Since,
$k_{i}$ = 48
$k_{f}$ = 48

Therefore,
48=48.

Ans-3) Hence proved: The gas IS obeying the Boyle's law.

-i

4 0

3 years ago

Read 2 more answers