What causes the

6. If you have a bottle with Helium & Nitrogen at room temperature, how does the speed of the particles compare?

Anon25 [30]

I believe it is C, as helium is one of the lightest noble gases making the particles move faster.

5 0

4 years ago

The equilibrium constant for the gas phase reaction

stellarik [79]

The equilibrium constant is found by [product]/[reactant]

If the equilibrium constant is very small, such as 4.20 * 10^-31, then that means at equilibrium there is very little product and a lot of reactant.
And likewise, if there is a lot of product formed, and very little reactant, then the K value will be very large, which tells us that it is predominantly product.

At equilibrium, for any reaction, there will always be some reactant and some product present. There cannot be zero reactant or zero product. Also keep in mind that the equilibrium constant is dependent on temperature.

At equilibrium, for your reaction, it is predominantly reactants.

6 0

3 years ago

The daily production of carbon dioxide from an 780.0 mw coal-fired power plant is estimated to be 3.3480 x 104 tons (not metric)

valkas [14]

The production of $CO_{2}$ is $3.3480\times 10^{4} tons/day$ . Converting mass into kg,

1 ton=907.185 kg, thus,

$3.3480\times 10^{4} tons=3.037\times 10^{7} kg$

Thus, production of $CO_{2}$ will be $3.037\times 10^{7} kg/ day$ .

The specific volume of $CO_{2}$ is $0.0120 m^{3}/kg$ .

Volume of $CO_{2}$ produced per day can be calculated as:

$V=Specific volume\times mass$

Putting the values,

$V=0.0120 m^{3}/kg\times 3.037\times 10^{7} kg=364440 m^{3}/day$

Thus, volume of $CO_{2}$ produced per year will be:

$V=\frac{365 days}{1 year}(364440 m^{3}/day)=1.33\times 10^{8}m^{3}/year$

Thus, in 4 year volume of $CO_{2}$ produced will be:

$V=1.33\times 10^{8}m^{3}/year\times 4 years=5.32\times 10^{8}m^{3}$

8 0

3 years ago

SO

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

Which cell organelle is like the construction site, where proteins are built? Which cell organelle packages and transports the p

german

Explanation:

The endoplasmic reticulum consists of a network of a tube-like passageway through which proteins from the ribosomes are able to be moved within a cell as the road system allows for movement throughout the city.

7 0

3 years ago