Ask question

Ask question

All categories

2 years ago

9

A physical change is ...

1 answer:

AnnZ [28]2 years ago

5 0

Answer:

the answer is c I just took the test

You might be interested in

A gas is collected at 30 Celsius and has a pressure of 200 kPa. What pressure would it exert if the temperature changed to 40 Ce

garik1379 [7]

Answer:

$206.6\ \text{kPa}$

Explanation:

$P_1$ = Initial pressure = 200 kPa

$P_2$ = Final pressure

$T_1$ = Initial temperature = $30^{\circ}\text{C}$

$T_2$ = Final temperature = $40^{\circ}\text{C}$

We have the relation

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}\\\Rightarrow P_2=\dfrac{T_2}{T_1}P_1\\\Rightarrow P_2=\dfrac{40+273.15}{30+273.15}\times 200\\\Rightarrow P_2=206.6\ \text{kPa}$

The pressure that would be exerted after the temperature change is $206.6\ \text{kPa}$ .

6 0

2 years ago

The cell theory was construted by

NeTakaya

Answer:

Matthias Jakob Schleiden

Explanation:

The cell was initially discovered by Robert Hooke, but Schleiden developed the theory.

Hope this helps! :)

4 0

3 years ago

Aluminum reacts with iron nitrate to form aluminum nitrate and iron. What is the

Lelechka [254]

The reactants are aluminum and iron nitrate.

5 0

2 years ago

Explain why vegetables last longer in a fridge.

jeka57 [31]

It has to do with the releasing of ethylene, which speeds up the ripening process

6 0

2 years ago

Read 2 more answers

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

andre [41]

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

$K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

$K_{p} = K_{c}RT$

$0.003691 = K_{c} \times 8.314 \times 1500$

$K_{c} = 2.9 \times 10^{-7}$

Also, $K_{c} = \frac{K_{f}}{K_{r}}$

or, $K_{f} = K_{c} \times K_{r}$

= $2.9 \times 10^{-7} \times 400.613$

= $1.1617 \times 10^{-4} m^{6}/mol^{2}s$

Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$ .

6 0

3 years ago