Ask question

Ask question

All categories

VashaNatasha [74]

2 years ago

9

(Hurry ASAP!!!) There are 1.5 moles of methane (CH4) in a tank. It has a pressure of 4.4 atm at 173°C. Find the volume occupied

by the gas.

1 answer:

slamgirl [31]2 years ago

8 0

Answer:

hope this help !

Explanation:

Use the given functions to set up and simplify 173 ° C .

1.5 =

CH4 = CH4

4.4 = CH4

173 ° C = CH4

You might be interested in

Write the cell notation for the voltaic cell that incorporates each of the following redox reactions:(a) Al(s) + Cr³⁺(aq) → Al³⁺

Sladkaya [172]

The cell notation for the given voltaic cell is as follows

Al(s) | Al³⁺(aq) || Cr³⁺ (aq) | Cr(s)

<h3>What is a voltaic cell?</h3>

The device in which electrons are transferred through an external source is known as a voltaic cell.

The electron flow of a Voltaic cell allows redox reactions to produce energy in the form of electricity.

It converts the energy of spontaneous redox reaction into electrical energy

An anode, a cathode, and a salt bridge are all parts of a voltaic cell's setup. The salt bridge serves to neutralize the system.

To write the cell notation, identify the oxidation and reduction reactions.

At the anode, oxidation takes place

$Al(s) \longrightarrow Al^{3+} (aq) + 3e^-$

At the cathode, reduction takes place

$Cr^{3+} (aq) + 3e^- \longrightarrow Cr(s)$

While writing cell notation, the anode is written on the left, and the cathode is written on the right

A single vertical line differentiates between the phases and a double vertical line represents the salt bridge.

Now, write the cell notation of the given Voltaic cell

Al(s) | Al³⁺(aq) || Cr³⁺ (aq) | Cr(s)

Learn more about Voltaic cells:

brainly.com/question/14312871

#SPJ4

7 0

2 years ago

When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

3 0

3 years ago

A popular car has a gas tank that holds a maximum of 14.5 gallons of fuel. What is

gulaghasi [49]

Taking into account the change of units, the size of the tank is 54.88 liters.

In first place, the rule of three is a tool used to quickly solve problems involving a proportional relationship between two variables.

The rule of three is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other), the direct rule of three should be applied.

The rule of three is applied as follows, with A, B and C being known values and D being an unknown value: if A maintains a certain relationship with B, and it is known that C and D have the same relationship, it is possible to calculate the unknown value D using the following expression:

$D=\frac{CxB}{A}$

In this case, the rule of three can be applied as follows: if 1 gallon is equal to 3.785 liters, 14.5 gallons are equal to how many liters?

$D=\frac{14.5 gallonsx3.785 liters}{1 gallon}$

Solving:

D= 54.88 liters

In summary, the size of the tank is 54.88 liters.

Learn more with this example:

7 0

2 years ago

Read 2 more answers

The human eye has an osmotic pressure of 8. 00 atm at 37. 0 °c. what concentration (in moles/l) of a saline (nacl) solution will

Gelneren [198K]

0.3147 concentration (in moles/l) of a saline (NaCl) solution will provide an isotonic eyedrop solution.

Isotonic eye drops

Because it might result in eye discomfort or tissue damage if it is not maintained, isotonicity is regarded as a crucial component of ophthalmic medicines. A few drops of blood are mixed with the test preparation before being examined and judged under a microscope at a magnification of 40. Isotonic solutions are those that have the same amount of water and other solutes in them as the cytoplasm of a cell. Since there is no net gain or loss of water, placing cells in an isotonic solution will not cause them to either shrink or swell.

We can calculate the osmotic pressure exerted by a solution using the following expression.

π = M . R . T

where,

π is the osmotic pressure

M is the molar concentration of the solution

R is the ideal gas constant

T is the absolute temperature

The absolute temperature is 37 + 273 = 310 K

π = M . R . T

8 = (X mol/L) . (0.082atm.L/mol.K) . 310 K = 0.3147 mol/L

To learn more about osmotic pressure refer:

brainly.com/question/5041899

#SPJ4

7 0

2 years ago

CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?

Kipish [7]

<h3>Answer:</h3>

64 g O₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 36 g H₂O

[Solve] x g O₂

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
Divide/Multiply [Cancel Units]: $\displaystyle 63.929 \ g \ O_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

63.929 g O₂ ≈ 64 g O₂

8 0

3 years ago

Read 2 more answers