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3 years ago

6

What does it mean when your teacher wants you to round 143 to two significant figures

1 answer:

zhenek [66]3 years ago

5 0

Answer:The answer is 3...I think

Explanation:

143

Sig Figs

3

Decimals

0

Scientific Notation

1.43 × 102

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What are examples of everyday chemical reactions?

Softa [21]

<span>Photosynthesis Is a Reaction To Make Food</span>

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3 years ago

The heat of vaporization for benzaldehyde is 48.8 kj/mol, and its normal boiling point is 451.0 k. use this information to deter

user100 [1]

Answer:

The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

Explanation:

To solve this problem, we use Clausius Clapeyron equation: ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂).
The first case: P₁ = 1 atm = 760 torr and T₁ = 451.0 K.
The second case: P₂ = <em>??? needed to be calculated</em> and T₂ = 61.5 °C = 334.5 K.
ΔHvap = 48.8 KJ/mole = 48.8 x 10³ J/mole and R = 8.314 J/mole.K.
Now, ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂)
ln(760 torr /P₂) = (48.8 x 10³ J/mole / 8.314 J/mole.K) (1/451 K - 1/334.5 K)

ln(760 torr /P₂) = (5869.62) (-7.722 x 10⁻⁴) = -4.53.

(760 torr /P₂) = 0.01075
Then, P₂ = (760 torr) / (0.01075) = 70691.73 torr.

So, The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

7 0

3 years ago

The nucleus of an isotope of sulfur contains 34 neutrons. what is the atomic number of this isotope?

Flauer [41]

The atomic number of an element never changes no matter how many neutrons they have so Sulfur would just have an atomic number of 16 like normal.

8 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!

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3 years ago

A firework exploding is an example of a (4 points) physical change chemical change phase change nuclear change

Irina-Kira [14]

A firework exploding is a chemical change as soon as you light the fuse the combustion reacts with the gunpowder.

I hope this helps, good luck!!!!

5 0

3 years ago