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3 years ago

What happens when sodium and sulfur combine

Chemistry

2 answers:

german3 years ago

6 0

Answer:

It emits hydrogen sulfide...smells like rotten eggs..

ty:)pls let me know whether this is ryt:D

Mila [183]3 years ago

6 0

Answer:

Hey!

Your answer is that it creates hydrogen sulfide gas / liquid...

Which really smells...

Explanation:

HOPE THIS HELPS!

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Describe two ways the material in this lesson

IgorLugansk [536]

Answer:

- Dalton used creativity to modify Proust's experiment and interpret the results.

- Thomson used creativity to interpret the results of the cathode ray tube experiment.

Explanation:

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3 years ago

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Explain how Newton’s cradle demonstrates the law of conservation of energy.

UNO [17]

When power goes up it must come down.

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3 years ago

Calculate the percent composition by mass of oxygen in HClO

r-ruslan [8.4K]

Answer:

percentage composition of O in HClO IS 30.47

Explanation:

percentage composition of oxygen in HClO =(atomic mass of O)/(atomic mass of HClO)×100

= $\frac{16}{1+35.5+16}\times 100$

=30.47

4 0

3 years ago

Identify each pair of elements as likely to participate in ionic bonding when bonded together. Use your periodic table.

alekssr [168]

Answer:

1) NaCl, 5)MgS

Explanation:

What makes something an Ionic Bond:

Ionic bonds are generally between two oppositely charged ions. Usually salts are formed.
A metal and nonmetal are also generally considered to be ionic.
Electronegativity also plays a role in a difference of electronegativity of elements being greater than 2.0.

For 1) Na has a charge of 1+ and Cl has a charge of 1-. They both come together to form table salt. Their electronegativity is also larger than 2.0. Also Na is a metal and Cl ion is a nonmetal.

For 2) C and O are both nonmetals and are covalent bonds

For 3) Water is a polar covalent molecule because there is a electronegativity difference between H and O, but not great. H and O are also nonmetals.

For 4) While Fe is a metal and O is a nonmetal, their electronegativity difference is not greater than 2.0. This makes it polar covalent.

For 5) Mg is a metal and S is a nonmetal. They have a large electronegativity difference. Also Mg has a charge of 2+ while S has a charge of 2-.

5 0

3 years ago

The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.350 atm? (∆Hvap = 28.5

victus00 [196]

Answer:

The temperature at which the vapor pressure would be 0.350 atm is 201.37°C

Explanation:

The relationship between variables in equilibrium between phases of one component system e.g liquid and vapor, solid and vapor , solid and liquid can be obtained from a thermodynamic relationship called Clapeyron equation.

Clausius- Clapeyron Equation can be put in a more convenient form applicable to vaporization and sublimation equilibria in which one of the two phases is gaseous.

The equation for Clausius- Clapeyron Equation can be expressed as:

$\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }$

where ;

$P_1$ is the vapor pressure at temperature 1

$P_ 2$ is the vapor pressure at temperature 2

∆Hvap = enthalpy of vaporization

R = universal gas constant

Given that:

$P_1$ = 1 atm

$P_ 2$ = 0.350 atm

∆Hvap = 28.5 kJ/mol = 28.5 × 10³ J/mol

$T_1$ = 282 °C = (282 + 273) K = 555 K

R = 8.314 J/mol/k

Substituting the above values into the Clausius - Clapeyron equation, we have:

$\mathtt{In \dfrac{P_2}{P_1}= \dfrac{\Delta \ H _{vap}}{R} \begin {pmatrix} \dfrac{T_2 -T_1}{T_2 \ T_1} \end {pmatrix} }$

$\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{T_2 - 555}{555T_2} \end {pmatrix} }$

$\mathtt{In \begin {pmatrix} \dfrac{0.350}{1} \end {pmatrix} } = \dfrac{28.5 \times 10^3 }{ 8.314 } \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$- 1.0498= 3427.953 \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$\dfrac{- 1.0498}{3427.953}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$- 3.06246906 \times 10^{-4}= \begin {pmatrix} \dfrac{1}{555}- \dfrac{1}{T_2} \end {pmatrix} }$

$\dfrac{1}{T_2} = \begin {pmatrix} \dfrac{1}{555}+ (3.06246906 \times 10^{-4} ) \end {pmatrix} }$

$\dfrac{1}{T_2} = 0.002108048708$

$T_2 = \dfrac{1}{0.002108048708}$

$\mathbf{T_2 }$ = 474.37 K

To °C ; we have $\mathbf{T_2 }$ = (474.37 - 273)°C

$\mathbf{T_2 }$ = 201.37 °C

Thus, the temperature at which the vapor pressure would be 0.350 atm is 201.37 °C

3 0

3 years ago