All categories

2 years ago

Connect and Apply scientific knowledge to explain the following situations:

Chemistry

1 answer:

mixer [17]2 years ago

7 0

Answer:

b.Rearrange the two equations to make each variable the focus (you will end up with three variations for each equation).

You might be interested in

How many dimes could you trade for 360 pesos? $1 = 1500 pesos.

sveta [45]

Answer:

The correct option is (d).

Explanation:

It is given that,

1$ = 1500 pesos

We need to convert 360 pesos into dimes

We can convert 360 pesos to dollars as follows:

$360\ \text{pesos}=\$\dfrac{1}{1500}\times 360\\\\=$0.24$

360 pesos is equal to $0.24

Also, 1 dollar = 10 dimes

We can covert 0.24 dollar to dimes as follows :

0.24 dollar = 10 × 0.24 dimes

0.24 dollar = 2.4 dimes

360 pesos = 2.4 dimes

7 0

2 years ago

If 5,800 j of energy are applied to a 15.2 kg piece of lead, by how much does the temp change if the specific heat of lead is 0.

scoray [572]

Answer:

The answer will be 2.98K

Explanation:

Using the formula:

Q = mc∆T

Q= 5,800 (heat in joules)

m= convert 15.2kg to g which is 15200g (mass in grams)

c= 0.128 J/g °c (Specific heat capacity)

∆T= what we need to find (temperature change)

5800J = 15200g x 0.128 x ∆T

= 2.98K

7 0

3 years ago

Consider this equilibrium:

Ilia_Sergeevich [38]

Answer:

the answer is option E they are bronsted lowry acid

7 0

2 years ago

Need help answering

svp [43]

Answer:

Your answer would be E

Explanation:

4 0

2 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

2 years ago