If a material conducts heat easily, its a good _____. It has to be a 18 letter word

Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What

Mariana [72]

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $CH_4$ :-

Mass of $CH_4$ = 1.28 g

Molar mass of $CH_4$ = 16.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.28\ g}{16.04\ g/mol}$

$Moles_{CH_4}= 0.0798\ mol$

For $O_2$ :-

Mass of $O_2$ = 10.1 g

Molar mass of $O_2$ = 31.998 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{10.1\ g}{31.998\ g/mol}$

$Moles_{O_2}= 0.3156\ mol$

According to the given reaction:

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

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

Limiting reagent is the one which is present in small amount. Thus, $CH_4$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0798\ moles= \frac{Mass}{44.01\ g/mol}$

Mass of $CO_2$ = 3.51 g

Theoretical yield = 3.51 g

3 0

3 years ago

What part of an Atom undergoes change during radioactive decay

Aloiza [94]

It is clear that the core is subject to change

7 0

3 years ago

A solution of which of the following coordination compounds will form a precipitate when treated with aqueous AgNO3?

lidiya [134]

Answer:

[Cr(NH3)6.]C13

Explanation:

Alfred Werner's coordination theory (1893) recognized two kinds of valency;

Primary valency which are nondirectional and secondary valency which are directional.

Hence, the number of counter ions precipitated from a complex depends on the primary valency of the central metal ion in the complex.

We must note that it is only these counter ions that occur outside the coordination sphere that can be precipitated by AgNO3.

If we consider the options carefully, only [Cr(NH3)6.]C13 possess counter ions outside the coordination sphere which can be precipitated when treated with aqueous AgNO3.

3 0

3 years ago

Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.

BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = 240.9

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = 144

Now we can find percentage composition / percentage by mass of oxygen.

% composition = $\frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound}$ × 100

% composition = $\frac{144}{240.9}$ × 100 = 59.776%

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0

2 years ago

Please provide thorough explanation

Arlecino [84]

Possible products in this reaction are the products 2 and 3.

<h3>What is the Friedel Crafts reaction?</h3>

The Friedel Crafts reaction is one in which the electrophile is created by a Lewis acid reaction between the AlCl3 and the alkylhalide reactant .

Now we know that there is the possibility of two products in this reaction due to a resonance shift as such possible products in this reaction are the products 2 and 3.

Learn more about Friedel Crafts reaction:brainly.com/question/14993566?

#SPJ1

4 0

2 years ago