Please Help! <br><br> How many molecules are in a 4.1 mole sample of CO?

Natalka [10]

Answer : first opinion and also last
Expiation : Note that these last two reactions, and 2H + 2H → 4He + γ, .Nuclear fusion is a reaction in which two nuclei are combined to form a larger nucleus. Nuclear fusion is a reaction in which two nuclei are combined, or fused, to form a larger nucleus. We know that all nuclei have less mass than the sum of the masses of the protons and neutrons that form them. The missing mass times c2 equals the binding energy of the nucleus—the greater the binding energy, the greater the missing mass.

5 0

2 years ago

What is the democritus findings?

Volgvan

nvm

4 0

2 years ago

Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the sys

8090 [49]

Answer:

Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.

Explanation:

The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:

$Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}$ .

Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both $[\mathrm{SO_2\, (g)}]$ and $[\mathrm{O_2\, (g)}]$ will increase if the pressure is increased through compression. However, because $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient $Q$ .

As a result, the increase in pressure will have no impact on the value of $Q\!$ . If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.

8 0

2 years ago

Gerald's science teacher mixed liquid X and liquid Y, both at room-temperature, in a large beaker. two beakers of clear solution

Pachacha [2.7K]

Answer:

the answer is C

Explanation:

7 0

2 years ago

Read 2 more answers

(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

Taya2010 [7]

Answer:

Here's what I get

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

$Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}$

3. Plot the initial points

The graph below shows the initial concentrations plotted on the vertical axis.

7 0

3 years ago

Please Help! How many molecules are in a 4.1 mole sample of CO?