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3 years ago

Which type of reaction is Mg + S → MgS?

Chemistry

2 answers:

Anestetic [448]3 years ago

6 0

Answer: Synthesis reaction

Explanation:

Synthesis reaction as refers to as direct combination is a reaction in which two or more chemical species combine to form more complex product.

neonofarm [45]3 years ago

5 0

Answer:

the answer is the second option

Explanation:

hope this helps

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combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

3 years ago

In molecular oxygen (O=O) which atom is partially positive?

saveliy_v [14]

the oxygen atom

Explanation:

Water is a molecular compound consisting of polar molecules that have a bent shape. The oxygen atom acquires a partial negative charge while the hydrogen atom acquires a partial positive charge.

4 0

3 years ago

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Suppose you put a whole antacid tablet in one glass of water and a crushed antacid tablet in another glass containing the same a

iogann1982 [59]

The crushed tablets would stop bubbling/fuzzing first because it has a smaller surface area which means that it would dissolve before the uncrushed tablets which has a larger surface area.

3 0

3 years ago

It takes 281.7 kJ of energy to remove 1 mole of electrons from the atoms on the surface of lithium metal. If lithium metal is ir

Alex17521 [72]

Answer:

The maximum kinetic energy of electron is = 2.93 × $10^{-19}$ Joule

Explanation:

We know that total energy

$E = \frac{hc}{\lambda}$ ------------ (1)

Here h = plank's constant = 6.62 × $10^{-34}$ J s

c = speed of light = 3 × $10^{8}$ $\frac{m}{s}$

$\lambda$ = 261 nm = 261 × $10^{-9}$ m

Put all these values in equation (1) we get

E = 7.6 × $10^{-19}$ J

We know that

Total energy = Energy to remove an electron + K.E of electron

Energy to remove an electron = $\frac{281.7 (1000)}{(6.023)10^{23} }$

Energy to remove an electron = 4.67 × $10^{-19}$ J

K.E of electron = Total energy - Energy to remove an electron

K.E of electron = 7.6 × $10^{-19}$ - 4.67 × $10^{-19}$

K.E of electron = 2.93 × $10^{-19}$ Joule

Therefore the maximum kinetic energy of electron is = 2.93 × $10^{-19}$ Joule

3 0

3 years ago

What is the enthalpy of combustion (per mole) of C4H10 (g)?

Artemon [7]

The balanced chemical reaction for the complete combustion of C4H10 is shown below:

C4H10 + (3/2)O2 --> 4CO2 + 5H2O

The enthalpy of formation are listed below:
C4H10: -2876.9 kJ/mol
O2: none (because it is pure substance)
CO2: -393.5 kJ/mol
H2O: -285.8 kJ/mol

The enthalpy of combustion is computed by subtracting the total enthalpy formation of the reactants from that of the products.

ΔHc = (4)(-393.5 kJ/mol) + (5)(-285.8 kJ/mol) - (-2876.9 kJ/mol)
= -<em>126.1 kJ</em>

Thus, the enthalpy of combustion of the carbon is -126.1 kJ.

5 0

3 years ago

Read 2 more answers