All categories

3 years ago

How many moles of nitrogen are needed to completely convert 6.34 mol of hydrogen?

Physics

2 answers:

Sergeu [11.5K]3 years ago

5 0

Answer:

n = 2.11 moles

Explanation:

As we know by the reaction of hydrogen and nitrogen

$N_2 + 3H_2 --> 2NH_3$

so here from above equation we know that for complete conversion of 3 moles of hydrogen we need 1 mole of nitrogen

so here we know that 6.34 moles of hydrogen is available

so we require 1/3 times of moles of nitrogen for complete reaction

so number of moles of hydrogen will be

$n = \frac{1}{3} \times 6.34$

$n = 2.11 moles$

so here we can say it requires 2.11 moles of nitrogen for complete reaction of 6.34 moles of hydrogen

____ [38]3 years ago

4 0

If we use the equation:
N2 + 3H2 --> 2NH3
Then
1 mol of Nitrogen required 3 moles of Hydrogen
x mols : 6.34mols
X = 6.34/3
X = 2.11 moles of Nitrogen are required.

You might be interested in

[01.02]

julia-pushkina [17]

That information describes the plane's speed.
If he had also told them what direction they were flying, then
they would have been able to put the two pieces of information
together, and they would know the plane's velocity.

6 0

3 years ago

Read 2 more answers

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=> $KE = \frac{ 1}{2} *F_g * r$

substituting values

$KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

$KE = 7.59 *10^{10} \ J$

7 0

3 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

3 years ago

A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.

wlad13 [49]

Answer:

12m/s

Explanation:

$v_f=v_o+at$

Let's call the velocity that the car maintains for 10 seconds $v_f_1$ , and the final velocity $v_f_2$ .

$v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s$

Hope this helps!

5 0

3 years ago

List 5 reason why water is not thermometric liquid

kolezko [41]

Answer:

(1) it is transparent so it makes the reading difficult. (2) it is volatile. (3) it is a poor conductor of heat. (4) it has a higher specific heat capacity, so it absorbs more heat from the body with which it is kept in contact. (5) Water cannot be used in thermometer because of its higher freezing point and lower boiling point than other liquids . If water is used in a thermometer , it will start phase change at 0oC and 100oC and will not measure temperature , out of this range .

PLEASE READ CAREFULLY AND PLEASE MARK AS BRAINLIEST :)

5 0

3 years ago