What is the name of this compound

What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?

katen-ka-za [31]

Answer:

D. N₂O

Explanation:

Let's assume we have 100 g of the compound. That means it consists of 63.61 grams of nitrogen and 36.69 grams of oxygen.

Converting masses to moles:

63.61 g N × (1 mol N / 14.01 g N) = 4.540 mol N

36.69 g O × (1 mol O / 16.00 g O) = 2.293 mol O

Normalize by dividing by the smallest:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

So there is approximately twice as many N atoms as O atoms. The empirical formula is therefore N₂O.

8 0

3 years ago

Cell notation will list each half-reaction:

likoan [24]

Answer:

on each side of the salt bridge, which is represented by a double vertical line

Explanation:

While writing a cell notation, the general convention is; anode || cathode. The anode and the cathode are separated by a double line. The anode is written on the lefthand side while the cathode is written on the righthand side.

The cell notation is a shorthand representation of a cell, hence any electrochemical cell can easily be produced based on its cell diagram.

6 0

3 years ago

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

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

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

<u> Theoretical yield = 3.51 g</u>

3 0

3 years ago

draw the structure of two acyclic compounds with 3 or more carbons which exhibits one singlet in the 1H-NMR spectrum

loris [4]

Answer:

attached below

Explanation:

Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum

a) Acetone CH₃COCH₃

Attached below is the structure

b) But-2-yne (CH₃C)₂

Attached below is the structure

5 0

3 years ago

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calcu

stealth61 [152]

Answer:

Explanation:

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)

Explanation;

According to Arrhenius equation:

i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)

Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K

T2 = 25 oC = (25 + 273) K = 298 K

i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)

i.e. ln(k2/0.000717) = -2.54738

i.e. k2/0.000717 = $e^{-2.54738}$

= 0.078286

Therefore, the required constant (k2) = 0.078286 * 0.000717 = $5.61*10^-^5$

6 0

3 years ago