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3 years ago

why do we heat the test tube in a glass beaker during a chemistry experiment and not directly over a flame?

1 answer:

7 0

It is not directly over a flame because it depends on the substance you might not want to heat it too much.you never know what could happen

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HELP ILL GIVE U FREE POINTS AND BRAINLEST !! Observe the plate boundary shown here. What type of plate boundary is this? Many ge

Studentka2010 [4]

The answer is B.
You can rule C out because divergent means moving away. Rule out A because there is an oceanic and continental plate, not 2 of the same type. Rule D out for the same reason.

3 0

3 years ago

Read 2 more answers

The energy that a log has is transformed when burned. How does the chemical energy of the log compare to the heat and light ener

SCORPION-xisa [38]

The amount of chemical energy is equal to the amount of heat and light energy.

This is due to the principle of conservation of energy, which states that the total energy of a system remains constant.

4 0

3 years ago

Which compound forms from 2 parts of hydrogen and 1 part of oxygen?

lina2011 [118]

Water, H2O, HOH these are all the same compound just worded differently

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3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

10. An atom that has 13 protons and 13 electrons is a neutral aluminum (Al) atom. An atom that has 13 protons and 10 electrons i

kotegsom [21]

Answer:

An atom of Al which has 13 protons and 10 electrons is Al cation (Al⁺³)

Explanation:

An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other.

For example,

Al atom has 13 protons and 13 electrons. The number of positive and negative charge is equal thus it will be neutral atom.

While the atom of Al which have 13 proton and 10 electron is not neutral. The positive charge is greater than negative by 3. Which means 3 electrons are lose by Al atom and form cation "Al⁺³".

Thus an atom of Al which has 13 protons and 10 electrons is Al cation (Al⁺³)

3 0

3 years ago