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arsen [322]
3 years ago
10

A graduated cylinder is filled with 20ml of water. A rock is dropped in to the graduated cyclinder and the volume of water rises

to 30ml
inside the graduated cylinder. The mass of the rock is 23g. What is the density of the rock?
2.3g/ml
0.77g/ml
5g/ml
1.15g/ml
Chemistry
1 answer:
Kitty [74]3 years ago
8 0

Answer:

\boxed {\boxed {\sf 2.3 \ g/mL}}

Explanation:

We are asked to find the density of a rock. Density is the substance's mass per unit volume. The formula for calculating density is:

\rho= \frac{m}{v}

The mass of the rock is 23 grams. The volume was found using water displacement. A known amount of water was measured, the rock was added to the graduated cylinder, and the water level was recorded again. The volume is the difference between the final and initial water level. The graduated cylinder originally had 20 milliliters and the water rose to 30 milliliters.

  • volume = final water level - initial water level
  • volume = 30 mL - 20 mL
  • volume = 10 mL

Now we know the mass and the volume.

  • m= 23 g
  • v= 10 mL

Substitute the values into the formula.

\rho= \frac{ 23 \ g }{30 \ mL}

Divide.

\rho= 2.3 \ g/mL

The density of the rock is <u>2.3 grams per milliliter.</u>

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the chloride of a metal M contains 47.25% of metal 1.0 gram of metal would be displaced from a compound by 0.88 gram of another
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The equivalent weight of M is approximately 31.8 g

The equivalent weight of N is approximately 27.98 g

Explanation:

The given parameters are;

The percentage of the the metal M in in the chloride = 47.25%

Where by the chemical formula for the metal chloride is MClₓ, we have;

47.25% of the mass of MClₓ = Mass of M = W

Therefore, we have;

\dfrac{0.4725}{W} = \dfrac{1}{W + 35.5 \cdot x}

0.4725 × (W +  35.5·x) = W

0.4725·W + 0.4725×35.5×x = W

W - 0.4725·W  = 16.77·x

0.5275·W = 16.77·x

W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M

The equivalent weight of M = 31.799 ≈ 31.8 g

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7 0
3 years ago
Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T
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Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

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Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

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PV=nRT

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P = pressure of the gas = 730.5 mmHg  

V = Volume of the gas = 9.49 L

T = Temperature of the gas = 20^oC=[20+273]K=293K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol

According to the reaction shown below as:-

2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when \frac{2}{3}\times 0.3794 moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

PV=nRT

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P = pressure of the gas = 735.5 mmHg  

V = Volume of the gas = 9.99 L

T = Temperature of the gas = 20^oC=[20+273]K=293K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol

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