What atom is bigger Boron or Carbon

The boiling process is considered to be____

Dmitry_Shevchenko [17]

your answer is D. physical change, because a new substance is not formed.

5 0

3 years ago

What process transform solar energy into chemical energy?

DIA [1.3K]

Green plants transform solar energy to chemical energy (mostly of oxygen) through the process known as photosynthesis, and electrical energy can be converted to chemical energy and vice versa through electrochemical reactions.

8 0

3 years ago

Covert from scientific notation 1.3 x 10¹²=

Lady_Fox [76]

Answer:

1,300,000,000,000

Explanation:

1.3 x 10^12

We want to convert this from scientific notation.

Tip: in scientific notation the exponent tells you how many place you move the decimal point over. If the exponent is negative you move the decimal point to the left. Ex. For, 4.1 x 10^-8, we would move the decimal point over 8 times to the left to get .00000041. When the exponent is positive we move over to the right. Ex. For, 7.6 x 10^7 we would move the decimal point over 7 times to the right to get 76,000,000

So to convert 1.3 x 10^12 we simply move over the decimal point over 12 times to the right.

1.3 x 10^12 ------> 1,300,000,000,000

Our answer is 1,300,000,000,000

6 0

3 years ago

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M