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2 years ago

Compare a sodium ion and neutral sodium atom in size. what causes the size difference

1 answer:

n200080 [17]2 years ago

4 0

A cation is smaller than its parent atom.
Therefore, here a sodium ion is smaller than the neutral sodium atom.

Reason: Since the sodium atom loses an electron to form an electropositive ion(cation), its nuclear charge increases(the number of protons exceeds the number of electrons) and the orbital electrons are pulled with a greater energy.
Thus,the sodium ion shrinks in size.

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Jlenok [28]

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2 years ago

A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and

Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = $99.7^oC$

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

$1atm=760mmHg$

or,

$1mmHg=\frac{1}{760}atm$

As, $1mmHg=\frac{1}{760}atm$

So, $752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm$

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

$K=273+^oC$

As, $K=273+^oC$

So, $K=273+99.7=372.7$

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

$(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole$

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

$\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}$

$\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole$

Thus, the mass of one mole of vapor is 32.0 g/mole

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2 years ago

Can we use electrical equipment near water? Explain Why?

tiny-mole [99]

Answer:

You may, but it is too risky.

Even though you are being cautious around using electric equipment around water, you'll never know what can happen. You might accidentally drop that piece of electrical equipment you are using into the water. Water can be splashed around by someone or something without you noticing it and it may affect the object you are using. Sometimes, if water comes in contact with an electrical object, it may cause you electric shocks or the equipment you are using has a chance of exploding and may hurt you. You can guarantee that waterproof electrical equipment is safe to use, but it is better not to risk it too much.

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2 years ago

I need help with problem 103 all of the questions.

AfilCa [17]

A. is decomposition so HCL = H2 + Cl2

not balanced cause hcl needs 2

2HCL = H2 + Cl2

balanced

b. Br2 + Al-i = AlBr3 + I2 single rep.

not balanced since br need 3 so watch carefully cause many changes needed

3Br2 + Al-i = AlBr3 + I2 not right is unbalanced so make it 2

3Br2 + Al-i = 2AlBr3 + I2 now left Al is unbal. so make 2 there

3Br2 + 2Ali = 2AlBr3 + I2

Balanced

C. Na + S = Na2S synthesis reaction is not bal. left Na needs 2

2Na + S = Na2S balanced.

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Answer:

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