<span>The student is incorrect because helium has 2 valence electrons and it's in group 18 because the first energy level is full. Although helium is placed in Group 18 which generally has 8 valence electrons, it does not have 8 valence electrons as the student suggested. It was grouped together with the noble gases because it exhibits similar properties with them. </span>

5 0

3 years ago

The rate constant for this second‑order reaction is 0.610 M − 1 ⋅ s − 1 0.610 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product

Darya [45]

Answer: It takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.

Explanation:

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

k = rate constant = $0.610M^{-1}s^{-1}$

$a_0$ = initial concentration = 0.860 M

a= concentration left after time t = 0.260 M

$\frac{1}{0.260}=0.860\times t+\frac{1}{0.860}$

$t=3.120s$

Thus it takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.

8 0

3 years ago

4. A student is doing experiments with CO2(q) . Originally a sample of the gas is in a rigid container at 299 Kand 0.70 atm. The

marta [7]

Answer:

b. 0,99atm

c. Answer is in the explanation

d. Answer is in the explanation

Explanation:

b. Using Gay-Lussac's law:

P₁T₂ = P₂T₁

P₁: 0,70 atm; T₂: 425K; P₂: ??; T₁: 299K

0,70atm×425K / 299K = <em>0,99 atm</em>

c. Using kinetic molecular theory, the increasing of temperature increases the kinetic energy of gas particles and if kinetic energy increases, the pressure increases. That means the increasing of temperature increases the pressure in the system.

d. Now, the increases in kinetic energy of gases increase the collisions betwen particles. As these intermolecular forces that are not taken into account in ideal gas law, the observed pressure will be different to the pressure predicted by ideal gas law.

I hope it helps!

8 0

2 years ago

Burka [1]

Answer:

$\large \boxed{\text{E) 721 K; B) 86.7 g}}$

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}$

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

$\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}$

7 0

3 years ago