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kvv77 [185]
1 year ago
12

A 92.8 gram sample of CuSO4•5H2O.(copper sulfate pentahydrate) is heated until water is released. How many grams of water were r

eleased?
Chemistry
2 answers:
Ivan1 year ago
7 0

Mass of water released =

92.8 g CuSO_{4}.5H_{2}O×\frac{5 * 18 g H_{2}O}{249.68 g CuSO_{4}.5H_{2}O}

= 33.45 g H_{2}O

babymother [125]1 year ago
5 0

<u>Answer:</u> The mass of water released from the given amount of copper sulfate is 33.46 grams.

<u>Explanation:</u>

We are given a compound having chemical formula CuSO_4.5H_2O

The molar mass of this compound = [63.55+32+(4\times 16)+5(16+(2\times 1))]=249.55g/mol

Mass of water molecule = [16+(2\times 1)]=18g/mol

In 249.55 grams of the compound, (5\times 18)g of water molecule is present.

So, in 92.8 grams of the compound, \frac{5\times 18}{249.55}\times 92.8=33.46g of water molecule.

Hence, the mass of water released from the given amount of copper sulfate is 33.46 grams.

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You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number
irina1246 [14]

Answer: 9.11\times 10^{-31}kg

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus 9.11\times 10^{-31}kg has three significant figures

7 0
2 years ago
If 4.9 kg of CO2 are produced during a combustion reaction, how many molecules of CO2 would be produced?
solmaris [256]

Answer:

6.7 x 10²⁶molecules

Explanation:

Given parameters

Mass of CO₂  = 4.9kg  = 4900g

Unknown:

Number of molecules  = ?

Solution:

To find the number of molecules, we need to find the number of moles first.

 Number of moles  = \frac{mass}{molar mass}

          Molar mass of CO₂  = 12 + 2(16)  = 44g/mol

   Number of moles  = \frac{4900}{44}  = 111.36mole

A mole of substance is the quantity of substance that contains the avogadro's number of particles.

       1 mole  = 6.02 x 10²³molecules

     111.36 moles  =   111.36 x 6.02 x 10²³molecules   = 6.7 x 10²⁶molecules

5 0
2 years ago
A driver with a nearly empty fuel tank may say she is "running on fumes."
tensa zangetsu [6.8K]

Answer:

The farthest the vehicle could travel (if it gets 20.0 miles per gallon on liquid gasoline) is 1.62 miles.

Explanation:

The automobile gas tank has a volume capacity of 15 gallons which can be converted to liters: 15 × 3.7854 = 56.781 liters

We can find the moles of gasoline by using the ideal gas equation: PV = nRT.

Make n (number of moles) the subject of the formula: n = PV/RT, where:

P = 747 mmHg

V = 56.781 liters

R (universal gas constant) = 0.0821 liter·atm/mol·K

T = 25 ∘C = (273 + 25) K = 298 K

1 atm (in the unit of R) = 760 mmHg

Therefore n = 747 × 56.781/(0.0821 × 760 × 298) = 2.281 mol.

Given that the molar mass of the gasoline = 101 g/mol,

the mass of gasoline = n × molar mass of gasoline = 2.281 mol × 101 g/mol = 230.38 g

the density of the liquid gasoline = 0.75 g/mL

In order to calculate the distance the vehicle can travel, we have to calculate volume of gasoline available = mass of the liquid gasoline ÷ density of liquid gasoline

= 230.38 g ÷ 0.75 g/mL = 307.17 mL = 0.3071 liters = 0.3071 ÷ 3.7854 = 0.0811 gallons

since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by the car = gallons available × miles per gallon = 0.0811 × 20 = 1.62 miles.

6 0
2 years ago
A gas is know to have a volume of 7.81 liters when the pressure is 754 torr what would be the volume when the pressure is change
Elis [28]

Answer:

The volume when the pressure is changed to 1.23 atm and temperature is constant will be <u><em>6.3075 L</em></u>.

Explanation:

Pressure and volume are related by Boyle's law that says:

"The volume occupied by a certain gas mass at a constant temperature is inversely proportional to the pressure"

Boyle's law is expressed mathematically as:

P * V = k

where:

  • P: Pressure
  • V: Volume
  • k: Constant

Assuming a certain volume of gas V1 is at a pressure P1 at the beginning of the experiment. By varying the volume of gas to a new V2 value, then the pressure will change to P2, and the following will be true:

P1 * V1 = P2 * V2

In this case you have:

  • P1= 754 torr= 0.9921 atm (1 atm=760 torr)
  • V1= 7.82 L
  • P2=1.23 atm
  • V2=?

Replacing:

0.9921 atm*7.82 L=1.23 atm*V2

Resolving:

V2=\frac{0.9921 atm*7.82 L}{1.23 atm}

V2≅6.3075 L

<u><em>The volume when the pressure is changed to 1.23 atm and temperature is constant will be 6.3075 L.</em></u>

6 0
1 year ago
The vapor pressure of liquid iodomethane, CH3I, is 100. mm Hg at 266 K. A 0.453 g sample of liquid CH3I is placed in a closed, e
Elanso [62]

Answer:

a. Yes

b. 143.5 mmHg

Explanation:

The vapor pressure is the pressure of the vapor that is in equilibrium with the liquid. At a constant temperature, some molecules of the liquid will vaporize, and then will do pressure at the surface of the liquid.

If the pressure at the container is higher then the vapor pressure, the liquid will evaporate.

a. Let's calculate the pressure at the container by the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (62,364 mmHg.mL/mol.K), and T is the temperature.

The molar mass of CH₃I is 142 g/mol

n = mass/ molar mass

n = 0.453/142

n = 0.0032 mol

P*370 = 0.0032*62,364*266

370P = 53,084.24

P = 143.5 mmHg

So, all the liquid will evaporate.

b. Because all liquid evaporates, when the equilibrium is reached, the pressure is the gas pressure: 143.5 mmHg.

3 0
2 years ago
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