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PilotLPTM [1.2K]
2 years ago
9

If 14 moles of oxygen react with 14 moles of hydrogen to produce water, what is the

Chemistry
1 answer:
Leokris [45]2 years ago
7 0
Hydrogen is the limiting reactant because when doing a stoichiometry equation for the reactants, hydrogen will be used completely by having a smaller yield and oxygen will be excess (7 moles to be exact)

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Where is almost all the mass of an atom​
VashaNatasha [74]
The nucleus!! This is made up of protons and neutrons that each weigh about 1 amu.

Electrons are not found in the nucleus and weigh almost nothing so chemistry in school doesn’t bother with them :)
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3 years ago
What are the forces that hold sodium and chloride ions together?
Phantasy [73]
Sodium is a metal, Chloride is a non-metal.

Right off the bat, you know that in order for both of these atoms to achieve a full valence shell that the metal has to lose electrons, and the non-metal has to gain them.

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6 0
3 years ago
Choose all the answers that apply.
Anni [7]

Answer:

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3 0
2 years ago
. A large quantity of very dilute aqueous HCl solution is neutralized by the addition of the stoichiometric amount of a 10-mol-%
ad-work [718]

Answer:

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Explanation:

5 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°
miskamm [114]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)    \Delta H_1=157kJ   ( × 4)

(2) P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)     \Delta H_2=-1207kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1835 kJ.

6 0
3 years ago
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