A bird on a long migration flies 63 kilometers per hour for 2900 kilometers. How long does he fly?

When a 375 mL sample of nitrogen is kept at constant temperature, it has a pressure of 1.2 atmospheres. What pressure does it ex

kolezko [41]

Answer:

It exerts a pressure of 3.6 atm

Explanation:

This is a gas law problem. We are looking at volume and pressure with temperature being kept constant, thus, the gas law to use is Boyle’s law. It states that at a given constant temperature, the volume of a given mass of gas is inversely proportional to the pressure of the gas.

Mathematically; P1V1 = P2V2

Let’s identify the parameters according to the question.

P1 = 1.2 atm

V1 = 375mL

P2 = ?

v2 = 125mL

We arrange the equation to make room for P2 and this can be written as:

P2 = P1V1/V2

P2 = (1.2 * 375)/125

P2 = 3.6 atm

3 0

2 years ago

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At 1.01 bar, how many moles of CO2 are released by raising the temperature of 1 litre of water from 20∘C to 25∘C

Tatiana [17]

Answer: 0.0007 moles of $CO_2$ is released when temperature is raised.

Explanation:

To calculate the number of moles, we use the ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.01 bar

V = Volume of the gas = 1L

R = Gas constant = $0.08314\text{ L bar }mol^{-1}K^{-1}$

Number of moles when T = 20° C

Temperature of the gas = 20° C = (273 + 20)K = 293K

Putting values in above equation, we get:

$1.01bar\times 1L=n_1\times 0.0814\text{ L bar }mol^{-1}K^{-1}\times 293K\\n_1=0.04146moles$

Number of moles when T = 25° C

Temperature of the gas = 25° C = (273 + 25)K = 298K

Putting values in above equation, we get:

$1.01bar\times 1L=n_2\times 0.0814\text{ L bar }mol^{-1}K^{-1}\times 298K\\n_2=0.04076moles$

Number of moles released = $n_1-n_2=0.04146-0.04076=0.0007moles$

Hence, 0.0007 moles of $CO_2$ is released when temperature is raised from 20° C to 25° C

5 0

2 years ago

How many grams of NH3 are needed to provide the same number of molecules as in 0.35 g of SF6 ?

Strike441 [17]

Answer:

There are 0.041 g of NH₃ in the same number of molecules as in 0.35 g of SF₆.

Explanation:

Using the molar mass of the chemical formula SF₆ you can find the number of moles of molecules in 0.35 g of such substance. Then, using the molar mass of NH₃, you can find mass in grams corresponding to the same number of molecules.

1. Find the molar mass of SF₆:

Atom atomic mass number of atoms total mass in 1 mole

S 32.065 g/mol 1 32.065 g

F 18.998 g/mol 6 6 × 18.998 = 113.988 g

=====================

molar mass of SF₆ = 146.053 g/mol

2. Find the number of moles in 0.35 g of SF₆:

number of moles = mass in grams / molar mass

number of moles = 0.35 g / 146.053 g / mol = 0.0024 mol

3. Find the molar mass of NH₃:

Atom atomic mass number of atoms total mass 1 mole

N 14.007 g/mol 1 14.007 g

H 1.008 g/mol 3 3 × 1.008 g = 113.988 g

=====================

molar mass of NH₃ = 17.031 g/mol

4. Find the mass in 0.0024 mol of NH₃:

mass in grams = number of moles × molar mass

mass = 0.0024 mol × 17.031 g/mol ≈ 0.041 grams

5. Conclusion:

There are 0.041 g of NH₃ in the same number of molecules as in 0.35 g of SF₆.

6 0

2 years ago

Using the following data table and graph, calculate the average k from the data. Amount CO2 (mL) Amount of White Solid (g) y x 1

UkoKoshka [18]

The k is the proportionality constant of the reaction. Graphically, this is the slope of the graph. Since the graph is linear, then there is only 1 value of k. To calculate this, choose two random points in the line. Suppose we use (0.15,10) and (0.30,20), calculate for the slope.

Slope = k = (10 - 20)/(0.15 - 0.30) = 66.67 mL CO₂/g CaCO₃

8 0

2 years ago

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A 2.25-g sample of magnesium nitrate, mg(no3)2, contains __________ mol of this compound.

MatroZZZ [7]

Mg(No3)2 is calculated as follows

moles = mass/molar mass
the molar mass of Mg(NO3)2 is = 148 g/mol

moles is therefore= 2.25 g / 148 g/mol= 0.0152 moles

Mg(No3)2 contain 0.0152 moles of the compound

6 0

2 years ago