What is 1006 in scientific notation?

Name:_____________________________________________________ Date:___________ Period:_________ 3/23 - 3/27 Assignment 1: Gas Law P

crimeas [40]

Answer:

The answers are;

1. 8.2 liters

2. 1214.84 ml

3. 318.027 K

4. 4.00 l.

Explanation:

1. Boyle's law states that the volume of given mass of gas is inversely proportional to its pressure at constant temperature

that is

P₁·V₁ = P₂·V₂

Where:

P₁ = Initial pressure = 40.0 mm Hg

V₁ = Initial volume = 12.3 liters

P₂ = Final pressure = 60.0 mm Hg

V₂ = Final volume = Required

From P₁·V₁ = P₂·V₂, V₂ is given by

$V_2=\frac{P_1\cdot V_1}{P_2} = \frac{40.0 mm Hg\cdot 12.3 l}{60.0 mm Hg}$ = 8.2 l

The volume reduces to V₂ = 8.2 liters

2. Here Charles law states that

$\frac{T_1}{V_1} =\frac{T_2}{V_2}$

T₁ = Initial temperature = 27.0 °C = 300.15 K

V₁ = Initial volume = 900.0 mL

T₂ = Final temperature = 132.0 °C = 405.15 K

V₂ = Final volume = Required

Therefore $V_2 =\frac{T_2\cdot V_1}{T_1}$ = $\frac{405.15 K\times 900.0 mL}{300.15 K}$ = 1214.84 ml

V₂ = 1214.84 ml

3. Gay-Lussac's Law states that

$\frac{T_1}{P_1} =\frac{T_2}{P_2}$

Where:

P₁ = Initial pressure = 15.0 atmospheres

T₁ = Initial temperature = 25.0 °C = 298.15 K

P₂ = Final pressure = 16.0 atmospheres

T₂ = Final temperature = Required

∴ $T_2 = \frac{T_1\times P_2}{P_1}$

= $\frac{298.15 K\times 16.0atm}{15.0atm}$ = 318.027 K

T₂ = 318.027 K

4. Avogadro's law states that,

Equal volume of all gases at the same temperature and pressure contain equal number of molecules.

Therefore if 5.00 moles of gas occupies 2.00 l volume, then

1 moles will occupy 2.00/5 l volume and

10 moles will occupy 2.00/5 × 10 or 4.00 l volume.

6 0

3 years ago

⦁answer Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg. ⦁ Determine the molar mass of a gas that has a

STALIN [3.7K]

Answer:

The correct answers are:

- Krypton: density= 2.8 g/L

- Molar Mass= 63.99 g/mol

- Mass of O₂= 15.29 g

Explanation:

The general equation of an ideal gas is the folllowing:

P x V = n x R x T

Where: P= pressure (in atm), V= volume; n= number of moles, R= gas constant (0,082 L.atm/K.mol) and T= temperature (in K).

For krypton:

P= 671 mmHg = 0,882 atm

V= 478 ml x 1000 ml/1 L= 0,478 L

T= 47ºC= 320 K

MM= 83.8 g/mol (from Periodic Table, Kr is an inert gas so it is a monoatomic gas)

P x V = n x R x T

Since the number of moles of a compound can be calculated by dividing the mass of compound (m) into its molar mass (MM):

n= m/MM

We can replace the expression in the first equation to obtain:

$P x V= \frac{m}{MM} x R x T$

m/V= $\frac{P x MM}{R x T}$

Density (d) is equal to the mass per volume (m/V), so we can directly calculate the density:

d= m/V= \frac{P x MM}{R x T}=

= (0.882 atm x 83.8 g/mol)/(0.082 L.atm/K.mol x 320 K)

= 2.81 g/L

For the gas:

d= 2.18 g/L

T= 66ºC= 339 K

P= 720 mmHg= 0.947 atm

d= \frac{P x MM}{R x T}

⇒MM = $\frac{dx R x T}{P}$

= (2.18 g/L x 0.082 L.atm/K.mol x 339 K)/(0.947 atm)

= 63.99 g/mol ≅ 64 g/mol

For the O₂:

V= 5.60 L

P= 1.75 atm

T= 250 K

MM(O₂) = 2 x Atomic Mass O= 2 x 16 g/mol= 32 g/mol

We can use the second equation:

P x V= \frac{m}{MM} x R x T

⇒ m = $\frac{P x V x MM}{R x T}$ = (1.75 atm x 5.6 L x 32 g/mol)/(0.082 L.atm/K.mol x 250 K)

= 15.29 g≅ 16 g

4 0

3 years ago

Read 2 more answers

Given the reactions below, (1) Na(s) + H2O(l) → NaOH(s) + 1/2 H2(g) LaTeX: \Delta H^o_{rxn}Δ H r x n o = −146 kJ (2) Na2SO4(s) +

slamgirl [31]

Answer:

ΔrxnH = -580.5 kJ

Explanation:

To solve this question we are going to help ourselves with Hess´s law.

Basically the strategy here is to work in an algebraic way with the three first reactions so as to reprduce the desired equation when we add them together, paying particular attention to place the reactants and products in the order that they are in the desired equation.

Notice that in the 3rd reaction we have 2 mol Na₂O (s) which is a reactant but with a coefficient of one, so we will multiply this equation by 1/2-

The 2nd equation has Na₂SO₄ as a reactant and it is a product in our required equation, therefore we will reverse the 2nd . Note the coefficient is 1 so we do not need to multiply.

This leads to the first equation and since we need to cancel 2 NaOH, we will nedd to multiply by 2 the first one.

Taking 1/2 eq 3 + (-) eq 2 + 2 eq 1 should do it.

Na₂O (s) + H₂ (g) ⇒ 2 Na (s) + H₂O(l) ΔrxnHº = 259 / 2 kJ 1/2 eq3

+ 2NaOH(s) + SO₃(g) ⇒ Na₂SO₄ (s) + H₂O (l) ΔrxnHº = -418 kJ - eq 2

+ 2Na (s) + 2 H₂O (l) ⇒ 2 NaOH (s) + H₂ (g) ΔrxnHº = -146 x 2 2 eq 1

Na₂O (s) + SO₃ (g) ⇒ Na₂SO₄ (s) ΔrxnHº = 259/2 + (-418) + (-146) x 2 kJ

ΔrxnH = -580.5 kJ

4 0

3 years ago

Give an example of a case where the branches of natural science appeared overlap

ZanzabumX [31]

Much of biological studies are actually chemistry because of chemical changes that occur within organisms. Similarly, large parts of chemistry are governed by physics.

5 0

3 years ago

Butane reacts with oxygen to produce carbon dioxide and water

Wittaler [7]

What exactly are you looking for?
This is the balanced equation.
2 C4H10g + 13 O2g ---> 8 CO2g + 10 H2Og

4 0

3 years ago