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3 years ago

What happens when a sodium ion is attracted to a chlorine ion?

Chemistry

2 answers:

Jobisdone [24]3 years ago

7 0

They form an ionic compound.

Sliva [168]3 years ago

5 0

when a sodium ion is attracted to a chlorine ion they form an ionic compound

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A piece of an unknown substance weighing 124.0 grams is heated in boiling water to 100.0oc. when the substance is placed in a ca

MakcuM [25]

The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J

The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:

7560J= 124g * (100-26)* specific heat

specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C

3 0

4 years ago

ASAP!

Andrei [34K]

Answer:

8.3334%

Explanation:

You have two masses. To find the percent of sodium chloride in water by mass, you divide the mass of NaCl by water. First, make both units the same. Easiest is to convert kg into g. 1.5kg = 1500g

125g NaCl/1500g H2O = 0.0833333333 ==> 8.3334%

5 0

3 years ago

Discuss the shapes of s- orbitals, p-orbitals and d-orbitals.

timurjin [86]

The azimuthal quantum number (l) determines its orbital angular momentum and describes the shape of the orbital.

s-orbitals (for example 1s, 2s) are spherically symmetric around the nucleus of the atom.

p-orbitals are dumb-bell shaped. l = 0,1...n-1, when l = 1, that is p subshell.

d-orbitals are butterfly shaped.

3 0

3 years ago

A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of

jek_recluse [69]

<u>Answer:</u> The volume when the pressure and temperature has changed is $1.6\times 10^2mL$

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

Let us assume:

$P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K$

Putting values in above equation, we get:

$\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL$

Hence, the volume when the pressure and temperature has changed is $1.6\times 10^2mL$

5 0

3 years ago

Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of

lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

$U_{2}-U_{1}=Q-W$

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

$dw=Pdv$