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3 years ago

How many molecules of NH3 are produced from 4.72x10 negative 4 power g of H2?

1 answer:

7 0

<span>9.40x10^19 molecules.
The balanced equation for ammonia is:
N2 + 3H2 ==> 2NH3
So for every 3 moles of hydrogen gas, 2 moles of ammonia is produced. So let's calculate the molar mass of hydrogen and ammonia, starting with the respective atomic weights:
Atomic weight nitrogen = 14.0067
Atomic weight hydrogen = 1.00794

Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol
Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol

Moles H2 = 4.72 x 10^-4 g / 2.01588 g/mol = 2.34140921086573x10^-4 mol
Moles NH3 = 2.34140921086573x10^-4 mol * (2/3) = 1.56094x10^-4 mol

Now to convert from moles to molecules, just multiply by Avogadro's number: 1.56094x10^-4 * 6.0221409x10^23 = 9.400197448261x10^19
Rounding to 3 significant figures gives 9.40x10^19 molecules.</span>

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A certain first-order reaction 45% complete in 65seconds, determine the rate constant and the half life for the process

masha68 [24]

The rate constant : k = 9.2 x 10⁻³ s⁻¹

The half life : t1/2 = 75.3 s

<h3>Further explanation</h3>

Given

Reaction 45% complete in 65 s

Required

The rate constant and the half life

Solution

For first order ln[A]=−kt+ln[A]o

45% complete, 55% remains

A = 0.55

Ao = 1

Input the value :

ln A = -kt + ln Ao

ln 0.55 = -k.65 + ln 1

-0.598=-k.65

k = 9.2 x 10⁻³ s⁻¹

The half life :

t1/2 = (ln 2) / k

t1/2 = 0.693 : 9.2 x 10⁻³

t1/2 = 75.3 s

3 0

2 years ago

A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature

Aliun [14]

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

$v_0=500$
$\alpha =1.12\times 10^{-4}$
$\Delta \epsilon = 25$

Now,

Increase in volume will be:

⇒ $\Delta V = \alpha\times v_0\times \Delta \epsilon$

By putting the given values, we get

$=1.12\times 10^{-4}\times 500\times 25$

$=1.12\times 10^{-4}\times 12500$

$=1.4 \ mL$

8 0

3 years ago

A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To what temperature must the gas in the

Ivahew [28]

<u>Answer:</u> The temperature to which the gas in the syringe must be heated is 720.5 K

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=1.88atm\\V_1=285mL\\T_1=355K\\P_2=2.50atm\\V_2=435mL\\T_2=?K$

Putting values in above equation, we get:

$\frac{1.88atm\times 285mL}{355K}=\frac{2.50atm\times 435mL}{T_2}\\\\T_2=\frac{2.50\times 435\times 355}{1.88\times 285}=720.5K$

Hence, the temperature to which the gas in the syringe must be heated is 720.5 K

8 0

3 years ago

a gas that exerts a pressure of 215 torr in a container with a volume of 51.0 mL will exert a pressure of ? torr when transferre

zhannawk [14.2K]

To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:

P1V1 = P2V2,

Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:

(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr

The final pressure exerted by the gas would be 0.593 torr.

Hope this helps!

3 0

3 years ago

If you burn 29.4 g of hydrogen and produce 263 g of water, how much oxygen reacted?

vovikov84 [41]

The reaction between hydrogen and oxygen to form water is given as:

$H_2 + O_2 \rightarrow H_2O$