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3 years ago

Under ordinary conditions of temperature and pressure, the particles in a gas are

1 answer:

4 0

<span>In normal conditions gas particles remain very distant from each other. They rarely collide and are stable. When temperature increases the gas particles begin to move faster and collide more, reducing the distance. When pressure increases the gas particles also pick up kinetic speed and are also closer to each other.</span>

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Consider the balanced chemical reaction when phosphorus and iodine react to produce phosphorus triodide: 2 P(s) + 3 I2(g) → 2 PI

melamori03 [73]

Answer:

Percent yield of PI3 = 95.4%

Explanation:

This is the reaction:

2P (s) + 3I2 (g) > 2PI3 (g)

Let's determine the moles of iodine that has reacted.

58.6 g / 253.8 g/mol = 0.231 mol

Ratio is 3:2. Let's make a rule of three to state the moles produced at 100 % yield reaction.

3 moles of I2 react to make 2 moles of PI3

0.231 moles of I2 would make (0.231 .2) / 3 = 0.154 moles of PI3

As we have produced 0.147 moles let's determine the percent yield.

(Yield produced / Theoretical yield) . 100 > (0.147 / 0.154) . 100 = 95.4%

5 0

3 years ago

What did J. J. Thomson observe when he applied electric voltage to a cathode ray tube in his famous experiment?

Nataly_w [17]

The glass opposite to the negative electrode started to glow. Hence, option B is correct.

<h3>What is a cathode ray tube?</h3>

A cathode-ray tube (CRT) is a specialized vacuum tube in which images are produced when an electron beam strikes a phosphorescent surface.

J.J. Thomson, through his famous Cathode ray experiment, proved that all atoms contain small negatively charged particles known as electrons. In the experiment, he applied electric voltage across a cathode ray tube. a fluorescent material coating was done on the positive side. When the voltage was applied, the positive side has glowing dots.

Hence, option B is correct.

Learn more about the cathode ray tube here:

brainly.com/question/14409449

#SPJ1

6 0

2 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

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

3 0

3 years ago

11cm+11.38cm+500.55cm=

Vika [28.1K]

Answer:522.93 centimeters

Explanation:

I calculated the numbers

8 0

3 years ago

\"Moving down group 2A (Alkaline Earth Metals), which element has the largest first ionization energy?\" Is the answer to this:

lesya692 [45]

Ionization energy is the energy required to remove the outermost electron from one mole of gaseous atom to produce 1 mole of gaseous in to produce a charge of 1. The greater the ionization energy, the greater is the chance f the electron to be removed from the nucleus. In this casse, Radium has the largest ionization energy.

6 0

3 years ago