An object with more mass has more kinetic energy than an object with less mass, if both objects are moving _____

Which of these is an abiotic factor of an aquarium system?

Vinvika [58]

I need the answer too

3 0

3 years ago

I need this so badly!

Alla [95]

Answer:

See below

Explanation:

Use Ideal Gas Law

PV = n RT using R = .082057366 l-atm/k-mol

T must be in Kelvin

solve for 'n'

.92 * 1.6 = n * .082057366 * 287

n = .0625 moles

then the mole weight: .0625 * x = .314

x = mole weight = 5.025 gm

"Systems with either very <u>low pressures or high temperatures</u> enable real gases to be estimated as “ideal.” "

4 0

2 years ago

Read 2 more answers

A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture

Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

$n=\frac{PV}{RT}$

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

$n=(\frac{1.45*2.50}{0.0821*293})$

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide = $8.60g(\frac{1mol}{44g})$

moles of carbon dioxide = 0.195 mol

Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

0.195 = 0.453 - 2X

2X = 0.453 - 0.195

2X = 0.258

$X=\frac{0.258}{2}$

X = 0.129

So, there are 0.129 moles of methane in the mixture.

moles of propane = 0.151 - 0.129 = 0.022

mole fraction of methane = $\frac{moles of Methane}{total moles}$

mole fraction of methane = $\frac{0.129}{0.151}$

mole fraction of methane = 0.854

Hence, the mole fraction of methane gas in the original gas mixture is 0.854.

6 0

3 years ago

Respiration describes the process that living cells use to release energy by combining sugar and oxygen. The primary chemical ch

True [87]

Answer:

C the mass of the reactant molecules is the same as the mass of the product

5 0

3 years ago

A mass of 80 grams of Bromine would be

Elis [28]

I’m pretty sure it’s a

8 0

3 years ago

An object with more mass has more kinetic energy than an object with less mass, if both objects are moving ______.