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4 years ago

There are 3.20 x 10^22 atoms of copper in the outer shell of pennies. How many grams of

Chemistry

1 answer:

Vesnalui [34]4 years ago

3 0

Convert to moles first and then use the molar mass of copper to convert to grams

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A student sets up the following equation to convert a measurement. (The stands for a number the student is going to calculate.)

Crank

Answer:

0.090 J/(mmol·°C) × (1000 mmol/mole × 1 kJ/(1000 J)) = 0.090 kJ/mole

Explanation:

The unit of conversion from kilo-Joules to Joules is given as follows;

1000 Joules = 1 kilo-Joule

The unit of conversion from milimoles to moles is given as follows;

1000 milimoles = 1 Mole

Therefore, we have

The value of the given expression is 0.090 J/(mmol·°C) × 1000 mmol/mole × 1 kJ/(1000 J) = 0.090 kJ/mole

0.090 J/milimole = 0.09 kJ/mole.

3 0

3 years ago

Substances that are likely to disassociate in water

mr_godi [17]

Answer:

sugar,dirt,spit

Explanation:

5 0

3 years ago

PLEASE HELP

Art [367]

1)water has a great capacity to hold a moderate heat energy

3 0

3 years ago

4. Explain what you’ve learned about significant figures. How many significant figures are in the measurement 0.03050 kg?

nasty-shy [4]

There are 4 significant figures! Start counting after the first non-zero digit :)

Hope this helps.

3 0

3 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago