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STatiana [176]
3 years ago
15

Name the subatomic particles in nucleus

Chemistry
1 answer:
trapecia [35]3 years ago
7 0
Protons, neutrons..................................
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Temperature is inversely related to the average kinetic energy of a gas.
Temka [501]
Incorrect, temperature is directly proportional to the avg. KE of a gas.
6 0
3 years ago
A carbon molecule that has a different arrangement of atoms is known as a/an
maw [93]
The correct response I believe is A. Isomer. If a carbon molecule possess more than one arrangement of how its atoms can be arranged, those other arrangements are known as isomers.
3 0
3 years ago
Which processes must happen for sand to eventually become metamorphic rock? In one to two sentences, name and briefly describe t
trapecia [35]

Answer:

Sand may be deposited as sediments that become sedimentary rocks after hardening, or lithifying. Extreme burial pressure, rising temperature at depth, and a lot of time, will transform just about any rock to become a metamorphic rock.

Explanation:

4 0
3 years ago
How many grams of KCI can be dissolved in 63.5. g of water at 80
Masteriza [31]

Answer:

35.8 g

Explanation:

Step 1: Given data

Mass of water: 63.5 g

Step 2: Calculate how many grams of KCl can be dissolved in 63.5. g of water at 80 °C

Solubility is the maximum amount of solute that can be dissolved in 100 g of solute at a specified temperature. The solubility of KCl at 80 °C is 56.3 g%g, that is, we can dissolve up to 56.3 g of KCl in 100 g of water.

63.5 g Water × 56.3 g KCl/100 g Water = 35.8 g KCl

8 0
3 years ago
An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To
antoniya [11.8K]

Answer: The new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

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

The expression of K_p for above reaction follows:

K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}         ........(1)

We are given:

P_{PCl_5}=217.0torr

P_{PCl_3}=13.2torr

P_{Cl_2}=13.2torr

Putting values in above equation, we get:

K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24

Now we have to calculate the new partial pressure of Cl_2.

P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}

217.0torr+13.2torr+P_{Cl_2}=263.0torr

P_{Cl_2}=32.8torr

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of Cl_2.

Now, the equilibrium is shifting to the reactant side. The equation follows:

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

Initial:             13.2         32.8            217.0

At eqm:         13.2-x      32.8-x         217.0+x

Putting values in expression 1, we get:

1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38

Neglecting the 40.4 value of 'x'  because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of PCl_5 = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of PCl_3 = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of Cl_2 = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0
3 years ago
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