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olasank [31]
4 years ago
13

A waterbed mattress has the following dimensions: 210 cm X 160 cm X 25 cm. The density of water is 62.4lb/ft3. What is the weigh

t of the bed in pounds?
Chemistry
1 answer:
artcher [175]4 years ago
7 0

<u>Answer:</u> The weight of water bed in pounds is 1850.16 lb

<u>Explanation:</u>

To calculate the volume of cuboid, we use the equation:

V=lbh

where,

V = volume of cuboid

l = length of cuboid = 210 cm

b = breadth of cuboid = 160 cm

h = height of cuboid = 25 cm

Putting values in above equation, we get:

V=210\times 160\times 25=8.4\times 10^5cm^3=29.65ft^3      (Conversion factor:  1cm^3=3.53\times 10^{-5}ft^3 )

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

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

Density of waterbed = 62.4lb/ft^3

Volume of waterbed = 29.65ft^3

Putting values in above equation, we get:

62.4lb/ft^3=\frac{\text{Mass of waterbed}}{29.65ft^3}\\\\\text{Mass of waterbed}=(62.4lb/ft^3\times 29.65ft^3)=1850.16lb

Hence, the weight of water bed in pounds is 1850.16 lb

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A flashbulb of volume 2.70 mL contains O2(g) at a pressure of 2.30 atm and a temperature of 30.0 °C. How many grams of O2(g) doe
Sholpan [36]
P = 2.30 atm

Volume in liter = 2.70 mL / 1000 => 0.0027 L

Temperature in K = 30.0 + 273 => 303 K

R = 0.082 atm

molar mass O2 = 31.9988 g/mol

number of moles O2 :

P * V = n * R* T

2.30 * 0.0027 = n * 0.082 * 303

0.00621 = n * 24.846

n = 0.00621 / 24.846

n = 0.0002499 moles of O2

Mass of O2:

n = m / mm

0.0002499 = m / 31.9988

m = 0.0002499 * 31.9988

m = 0.008 g
3 0
4 years ago
A 25.0 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter
Ira Lisetskai [31]

Answer:

m H2O = 56 g

Explanation:

  • Q = mCΔT

∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:

⇒ - (mCΔT)Al = (mCΔT)H2O

∴ m Al = 25.0 g

∴ Mw Al = 26.981 g/mol

⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al

⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C

⇒ Q Al = 1327.64 J

∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C

⇒ mH2O = 55.722 g ≅ 56 g

5 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
Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

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

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

PV=nRT

where,

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

where,

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

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0
3 years ago
How much of the original amount of an isotope is present after a period of four half lives
vfiekz [6]

After 2 half-lives there will be 25% (1/4th) of the original isotope, and 75% (3/4 th) of the decay product

<h3>What is Half life period ?</h3>

A half life is a measurement of the slope of an exponential decay function.

It is also defined as, the time it takes to halve the concentration of something in a process.

Each half life you will have half of what you had at the beginning of a given half life.

Learn more about half life here ;

brainly.com/question/9654500

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7 0
2 years ago
The atomic mass of an imaginary element X is 30.7377 amu. Element X has three isotopes. One isotope has an isotopic mass of 30.5
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Answer:

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