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13. Classify as man-made or natural extinction: Massive impact from asteroid or comet.

NISA [10]

Answer:

Natural extinction

Explanation:

Man doesn't play a part in space objects causing extinction

4 0

3 years ago

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

Which metal does not form cations of differing charges?

valkas [14]

Transition metals

Most transition metals differ from the metals of Groups 1, 2, and 13 in that they are capable of forming more than one cation with different ionic charges. As an example, iron commonly forms two different ions

8 0

3 years ago

Which step do you use first to calculate the number of grams of FeCl2 produced when starting with 30.3 g of Fe in this reaction?

Lisa [10]

Answer:

I dont care I want pionts

5 0

3 years ago

If 0.40 mol of H2 and 0.15 mol of O2 were to react as completely as possible to produce H2O what mass of reactant would remain?

liberstina [14]

Answer : The mass of reactant $H_2$ remain would be, 0.20 grams.

Solution : Given,

Moles of $H_2$ = 0.40 mol

Moles of $O_2$ = 0.15 mol

Molar mass of $H_2$ = 2 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2H_2+O_2\rightarrow 2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $O_2$ react with 2 mole of $H_2$

So, 0.15 moles of $O_2$ react with $0.15\times 2=0.30$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

The moles of reactant $H_2$ remain = 0.40 - 0.30 = 0.10 mole

Now we have to calculate the mass of reactant $H_2$ remain.

$\text{ Mass of }H_2=\text{ Moles of }H_2\times \text{ Molar mass of }H_2$

$\text{ Mass of }H_2=(0.10moles)\times (2g/mole)=0.20g$

Therefore, the mass of reactant $H_2$ remain would be, 0.20 grams.

3 0

4 years ago