If 7.75 L of radon gas is at 1.55 atm and -19 °C, what is the volume at STP?

How many molecules are in 3.2 mol of ammonia,NH3

EastWind [94]

Here we are talking about ammonia which is a molecule. Now 1 mole of ammonia also contains 6.022<span> * 10^</span>23 molecules<span> of ammonia. In one molecule of ammonia,we have 1 nitrogen atom and 3 hydrogen atom.</span>

5 0

3 years ago

Which statement about the movement of Earth's crust along a fault line is true?

olga_2 [115]

The answer to your question is Different types of movement occur along different types of faults

7 0

3 years ago

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 1.74 g of butane is

xeze [42]

Answer : The mass of $CO_2$ produced will be, 5.3 grams.

Explanation : Given,

Mass of $C_4H_{10}$ = 1.74 g

Mass of $O_2$ = 11 g

Molar mass of $C_4H_{10}$ = 58 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $C_4H_{10}$ and $O_2$ .

$\text{Moles of }C_4H_{10}=\frac{\text{Mass of }C_4H_{10}}{\text{Molar mass of }C_4H_{10}}=\frac{1.74g}{58g/mole}=0.03moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{11g}{32g/mole}=0.34moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_4H_{10}$ react with 13 mole of $O_2$

So, 0.03 moles of $C_4H_{10}$ react with $\frac{13}{2}\times 0.03=0.195$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C_4H_{10}$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$ .

As, 2 moles of $C_4H_{10}$ react to give 8 moles of $CO_2$

So, 0.03 moles of $C_4H_{10}$ react to give $\frac{8}{2}\times 0.03=0.12$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$ .

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(0.12mole)\times (44g/mole)=5.3g$

Therefore, the mass of $CO_2$ produced will be, 5.3 grams.

5 0

3 years ago

PLEASE HELP!! I'LL MARK U BRAINIEST!! ALSO PLZ IF YOU CAN SHOW THE WORK

mel-nik [20]

Answer:

9.63×10²⁴ particles of H₂O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

From the balanced equation above,

1 mole of CH₄ reacted to produce 2 moles of H₂O.

Next, we shall determine the number of mole of H₂O produced by the reaction of 8 moles of CH₄. This can be obtained as follow:

From the balanced equation above,

1 mole of CH₄ reacted to produce 2 moles of H₂O.

Therefore, 8 moles of CH₄ will react to produce = 8 × 2 = 16 moles of H₂O.

Thus, 16 moles of H₂O were obtained from the reaction.

Finally, we shall determine the number of particles of H₂O produced. This can be obtained as follow:

Number of mole of H₂O produced = 16 moles

Number of particles produced =?

From Avogadro's hypothesis,

1 mole = 6.02×10²³ particles

Therefore,

16 mole = 16 mole × 6.02×10²³ particles / 1 mole

16 moles = 9.63×10²⁴ particles

Therefore, 9.63×10²⁴ particles of H₂O were produced.

6 0

3 years ago

Calculate the molar mass of BA(NO3)2

VikaD [51]

Hey there!

Ba(NO₃)₂

Ba: 1 x 137.327 = 137.327

N: 2 x 14.007 = 28.014

O: 6 x 16 = 96

-----------------------------------

261.341 g/mol

The molar mass of Ba(NO₃)₂ is 261.341 g/mol.

Hope this helps!

6 0

3 years ago