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adelina 88 [10]
3 years ago
6

How do I find the number of moles and molar mass of Gas A and Gas B with the information provided?

Chemistry
1 answer:
drek231 [11]3 years ago
3 0

Answer:

A.

Explanation:

hope this helped sorry if its wrong!

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2 years ago
What is the total number of molecules in 34.0?
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What is the element in which you are trying to find the number of molecules of?

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When the reaction A → B + C is studied, a plot 1/[A]t vs. time gives a straight line with a positive slope. What is the order of
zloy xaker [14]

Answer:

Second order

Explanation:

Given that:

When the reaction A → B + C is studied, a plot 1/[A]t vs. time gives a straight line with a positive slope.

From the integration method for the second order of reaction.

Suppose that:

rate = k₂[A]²

∴

\dfrac{1}{A}=\dfrac{1}{A_0}+k_2t

Therefore, a plot of the linear function \dfrac{1}{A} versus t will be linear with a positive slope k₂ and the intercept on the concentration axis will be \dfrac{1}{A_o}

The linear plot for a second order reaction can be seen in theimage attached below.

7 0
3 years ago
What do the toxic chemicals used in paving roads, rocket fuel, and rat poison have in common? What do the toxic chemicals used i
MArishka [77]

Answer: Arsenic

Explanation:. Arsenic,  is found in several different chemical forms in paving roads--- arsenic release from asphalt, rocket fuel--- from combustion of fuel causing release of arsenic and carbon monoxide  and among the composition of  rat poison in different oxidation states,which can lead to toxicity called  arsenic poisoning--- which occurs when much arsenic  is accumulated in the body causing  acute and chronic adverse health effects, from diarrhea to cancer.

3 0
3 years ago
Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given
tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

K_a= 1.34*10^{-5

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

\alpha = \sqrt{4.9629*10^{-5}}

\alpha = 7.044*10^{-3

percentage% (∝) = 7.044*10^{-3}*100

= 0.704%

For Part B; where Concentration of B = 7.84*10^{-2 M

\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }

\alpha = \sqrt{1.709*10^{-4} }

\alpha = 0.0130\\

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= 1.92*10^{-2} M

\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }

\alpha = \sqrt{6.802*10^{-4}}

\alpha =0.02608

percentage% (∝) = 0.02608  × 100%

= 2.60%

7 0
3 years ago
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