PLEASE HELP!!! The image represents the reaction between a certain number of molecules of N2 and H2.

Chemistry

2 answers:

KiRa [971]10 days ago

7 0

Response:

Two nitrogen molecules, N₂

Clarification:

The explanation aligns with the previous response. This was added for your convenience in accessing the correct answer.

lorasvet [956]10 days ago

4 0

Response:

The remaining reactant is nitrogen gas, N₂.

Clarification:

According to the information you've provided:

1. Left square: N₂(g)

3 units of two overlapping circles: this indicates 3 molecules of nitrogen gas, N₂(g).

2. Right square: H₂(g)

3 units of two overlapping circles: this represents 3 molecules of hydrogen gas, H₂(g).

3. Reaction overview

Assuming the reaction completes, producing the maximum amount of NH₃, you would follow the chemical equation:

N₂(g) + H₂(g) → NH₃(g)

This must be balanced:

N₂(g) + 3H₂(g) → 2NH₃(g)

This implies that 1 molecule (or 1 mole) of N₂(g) reacts with 3 molecules (or 3 moles) of H₂(g) to generate 2 molecules (or 2 moles) of NH₃(g).

Given that the squares indicate 3 molecules of each reactant, it becomes evident that 3 molecules of hydrogen gas can react with 1 molecule of nitrogen gas. Consequently, when this occurs, all hydrogen gas is consumed, resulting in 2 unreacted molecules of nitrogen gas. Thus, nitrogen gas remains as the leftover reactant.

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If the kinetic energy of a particle is equal to twice its rest mass, what is the velocity of the particle? Determine if relativi

lorasvet [956]

Answer:

The particle's velocity is calculated to be 2 m/s,

Explanation:

Kinetic energy refers to the energy an object possesses due to its movement. The formula for kinetic energy is:

$K.E=\frac{1}{2}mv^2$

Where:

m = the mass of the object

v = the object's velocity

A particle with mass m has a kinetic energy that is double its mass.

$K.E=2m$

$2m=\frac{1}{2}mv^2$

$v^2=\frac{4}{1}$

$v=2$

Since the velocity is measured in m/s, we determine that the particle's speed is 2 m/s.

8 0

8 days ago

What is [h3o ] in a solution of 0.25 m ch3co2h and 0.030 m nach3co2?

Anarel [852]

Hello!

To find [H₃O⁺], we will employ the Henderson-Hasselbalch equation, as this involves an acid and its conjugate base:

$pH=pKa+log( \frac{[A^{-}] }{[HA]} )$

$pH=4,76+log( \frac{0,030M}{0,25M} ) \\ \\ pH=3,84$

Next, we derive [H₃O⁺] using the definition of pH:

$pH=-log[H_3O^{+}]$

$[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M$

Hence, the concentration of [H₃O⁺] comes out to be 0.00014 M

Wishing you a good day!