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3 years ago

What the difference between a sodium atom (Na) and a sodium ion (Nal+)

Chemistry

2 answers:

vichka [17]3 years ago

7 0

Answer:

So a sodium atom has the electronic configuration 2,8,1 meaning it has 1 electron on its outer shell. It has a neutral charge since the number of electrons is equal to the number of protons.

A sodium ion is one that has lost the electron on its valence shell. The electronic configuration is 2,8 and it has a positive charge because it has more protons than electrons.

STALIN [3.7K]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf Number \ of \ electrons \ and \ net \ charge}}$

Explanation:

Remember that protons have a positive charge of +1 and electrons have a negative charge of -1. Atoms have protons and electrons.

A sodium atom has the electron configuration of 2-8-1. It has 11 electrons and its atomic number (number of protons) is also 11. This means it is a neutral atom because the charges of the electrons (-11) and protons (+11) balance each other out.

However, a sodium ion is different. It loses the 1 valence electron, so it only has 10 electrons. It still has the same atomic number of 11.

It is a not a neutral atom because the charges of the electrons (-10) and protons (+11) do not balance out. It has more protons and a net positive charge of +1. It is a cation or a positively charged ion.

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Is Ca + O --> CaO a balanced chemical equation? Is it obeying the law of conservation of matter? Question 5 options: Yes it i

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Answer is: Yes it is balanced. It has the same number of Ca and O on both the reactant side and the product side.

According to principle of mass conservation, number of atoms must be equal on both side of chemical reaction.

There is one calcium atom atom and one oxygen atom on both side of reaction, so chemical reaction is balanced.

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3 years ago

Predict the outcome of this reaction. Balancing doesn't matter. Mg + H2O -->

ELEN [110]

Answer:

MgO + H₂

Explanation:

The products of reaction is MgO and hydrogen gas.

This is a single displacement reaction.

In this reaction, the possibility of the reaction is determined by position of the atoms on the activity series.

Mg is higher than H on the activity series, so it is more reactive and it then displaces the hydrogen in water;

Mg + H₂O → MgO + H₂

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2 years ago

at atmoshperic pressure, a balloon contains 2.00L of nitrogen of gas. How would the volume change if the Kelvin temperature were

Nataly [62]

Answer: The percent change in volume will be 25 %

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=2L\\T_1=T_1\\V_2=?\\T_2=75\% \text{ of }T_1=0.75\times T_1$

Putting values in above equation, we get:

$\frac{2L}{T_1}=\frac{V_2}{0.75\times T_1}\\\\V_2=\frac{2\times 0.75\times T_1}{T_1}=1.5L$

Percent change of volume = $\frac{\text{Change in volume}}{\text{Initial volume}}\times 100$

Percent change of volume = $\frac{(2-1.5)}{2}\times 100=25\%$

Hence, the percent change in volume will be 25 %

5 0

3 years ago

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0

3 years ago

Complete combustion of 7.50 g of a hydrocarbon produced 23.1 g of CO2 and 10.6 g of H2O. What is the empirical formula for the h

r-ruslan [8.4K]

A CH compound is combusted to produce CO2 and H2O
CnHm + O2 -----> CO2 + H2O
Mass of CO2 = 23.1g
Mass of H2O = 10.6g
Calculate by mass of the compounds
For Carbon C, divide by molecular weight of CO2 and multiply with Carbon
molecular weight. So C in grams = 23.1 x (12.01 / 44.01) = 6.3 g C
For Hydrogen H, divide by molecular weight of H2O and multiply with Hydrogen molecular weight. So H in grams = 10.6 x (2.01 / 18.01) = 0.53 g C
= 1.18 of H
Calculate the moles for C and H
6.3 grams of C x (1 mole/12.01 g C) = 0.524 moles of C
1.18 grams of H x (1 mole/1.008 g H) = 1.17 moles of H
Divides by both mole entities with smallest
C = 0.524 / 0.524 = 1 x 4 = 4
H = 1.17 / 0.524 = 2.23 x 4 = 10
The empirical formula is C4H10.

7 0

3 years ago

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