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Artyom0805 [142]
2 years ago
9

Consider the reaction: AB As the reaction proceeds, the concentration of A drops from 0.625 M to 0.100 Min 11.6 seconds. What is

the average rate of the reaction (in M/s) during this time?
Chemistry
1 answer:
prisoha [69]2 years ago
3 0

Answer:

0.045 M/s

Explanation:

Given:

Initial concentration of A = 0.625 M

Final concentration of A = 0.100 M

Total time taken for the change of concentration = 11.6 seconds

The average rate of reaction is calculated as:

= \frac{\textup{Change in concentration of A}}{\textup{Time taken for the change}}

on substituting the respective values, we get

= \frac{0.625-0.100}{\textup{11.6}}

or

The average rate of reaction = 0.045 M/s

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The Top Thrill Dragster, a rollercoaster at Cedar Point in Ohio, soars through the air at 176ft/sec. How fast is this in mph?​
scoray [572]

Answer:

120mph

Explanation:

Google

divide the speed value by 1.467

or

176 times 60 second in a minute times 60 minutes in an hour

than divide by 5280 the amount of feet in a mile

7 0
2 years ago
Carrying capacity is _____.
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5 0
3 years ago
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Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if
mrs_skeptik [129]

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

  • <em>initial volume of gas in the container, V₁ = 2.65 L</em>
  • <em>initial number of moles of gas, n₁ = 0.12 mol</em>
  • <em>additional concentration, n = 0.182 mol</em>

The total number of moles of gas in the container is calculated as follows;

n_t = 0.12 + 0.182 = 0.302 \ mol

The final volume of gas in the container is calculated as follows;

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L

Thus, the final volume of the methane gas in the container is 6.67 L.

Learn more here:brainly.com/question/21912477

5 0
2 years ago
For many years chloroform (CHCl3) was used as an inhalation anesthetic in spite of the fact that it is also a toxic substance th
Debora [2.8K]

Hey there!:

Molar mass:

CHCl3 = ( 12.01 * 1 )+ (1.008 * 1 ) + ( 35.45 * 3 ) => 119.37 g/mol

C% =  ( atomic mass C / molar mass CHCl3 ) * 100

For C :

C % =  (12.01 / 119.37 ) * 100

C% = ( 0.1006 * 100 )

C% =  10.06 %

For H :

H% = ( atomic mass H / molar mass CHCl3 ) * 100

H% = ( 1.008 / 119.37 ) * 100

H% = 0.008444 * 100

H% = 0.8444 %

For Cl :

Cl % ( molar mass Cl3 / molar mass CHCl3 ):

Cl% =  ( 3 * 35.45 / 119.37 ) * 100

Cl% =  ( 106.35 / 119.37 ) * 100

Cl% = 0.8909 * 100

Cl% = 89.9%


Hope that helps!

4 0
3 years ago
Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container
Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

Explanation:

Here Pressure in a container is given as

P=\frac{1}{3} \rho

Here

  • P is the pressure which is to be calculated
  • ρ is the density of the gas which is to be calculated as below

                                         \rho =\frac{mass}{Volume of container}

        Here

                mass is to be calculated for 75 gas phase molecules as

                      m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times  10^{-21} g

              Volume of container is 0.5 lts

     So density is given as

                         \rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3

  • is the mean squared velocity which is given as

                                        =RMS^2

      Here RMS is the Root Mean Square speed given as 605 m/s so

                                      =RMS^2\\=(605)^2\\=366025

Substituting the values in the equation and solving

P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa

So the average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

6 0
3 years ago
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