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Oduvanchick [21]
3 years ago
5

For which one of the following is ΔHfo zero?

Chemistry
2 answers:
Vlada [557]3 years ago
6 0
The answer is most likely C
Sholpan [36]3 years ago
3 0
The standard enthalpy of formation is zero for B and D since by definition, the standard enthalpy of formation for an element at its standard state is zero. The monatomic oxygen is not the standard state of oxygen while carbon dioxide is already a molecule. 
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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179 x 1023 atoms of magnesium with 54.21 g of pho
wlad13 [49]
<h3>Answer:</h3>

18.58 liters of hydrogen gas

<h3>Explanation:</h3>

We are given;

  • The equation;

3Mg + 2H₃(PO₄) → Mg₃(PO₄)₂ + 3H₂

  • Atoms of Magnesium = 7.179 x 10^23 atoms
  • Mass of phosphoric acid as 54.21 g

We are required to determine the volume of hydrogen gas produced;

Step 1; moles of Magnesium

1 mole of an element contains 6.02 × 10^23 atoms

therefore;

Moles of Mg = (7.179 x 10^23 ) ÷ (6.02 × 10^23)

                   = 1.193 moles

Step 2: Moles of phosphoric acid

moles = Mass ÷ Molar mass

Molar mass of phosphoric acid = 97.994 g/mol

Therefore;

Moles of Phosphoric acid = 54.21 g ÷ 97.994 g/mol

                                           = 0.553 moles

Step 3: Determine the rate limiting reagent

From the mole ratio of Mg to Phosphoric acid (3 : 2);

1.193 moles of magnesium requires 0.795 moles of phosphoric acid while,

0.0553 moles of phosphoric acid requires 0.8295 moles of Mg

Therefore, phosphoric acid is the rate limiting reagent

step 4: Determine the moles of hydrogen produced

From the equation, w moles of phosphoric acid reacts to produce 3 moles of hydrogen;

Therefore; moles of Hydrogen = moles of phosphoric acid × 3/2

                                                   = 0.553 moles × 3/2

                                                   = 0.8295 moles

Step 5: Volume of hydrogen gas

1 mole of a gas occupies a volume of 22.4 liters at STP

Therefore;

Volume of Hydrogen = 0.8295 moles × 22.4 L/mol

                                  = 18.58 Liters

Therefore; 18.58 liters of hydrogen gas  will be produced

4 0
3 years ago
T/F___ At the eutectic composition, an alloy can solidify at a constant temperature.___ For effective dispersion strengthening,
azamat

Answer:

  • TRUE
  • FALSE
  • TRUE
  • FALSE
  • FALSE
  • TRUE

Explanation:

  • At the eutectic composition, an alloy can solidify at a constant temperature : TRUE . this is because at eutectic composition the type of reaction that takes place there is invariant reaction in its thermal equilibrium
  • For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round : FALSE. because the rounded shape will not cause a crack.
  • Intermetallic compounds are usually hard and brittle : TRUE. because Intermetallic compounds prevents dislocation movements and this makes them brittle and hard
  • For the effective dispersion and strengthening, the dispersed phase should be continuous : FALSE. this is because the dispersed precipitate must be small and not continuous
  • Stoichiometric intermetallic compounds exist over a range of compositions : FALSE
  • Faster solidification results in smaller interlamellar spacing : TRUE
4 0
3 years ago
Hi, I need help on the chemical names of the following two:
Lesechka [4]

Answer:

like how do you need help

3 0
3 years ago
Explain why metals are good conductors of electric current.
Phantasy [73]
For example, copper is used for electrical<span> wiring because it is a </span>good conductor of electricity<span>. </span>Metal<span> particles are held together by strong metallic bonds, which is why they have high melting and boiling points. The free electrons in </span>metals<span> can move through the </span>metal<span>, allowing </span>metals<span> to conduct </span>electricity<span>.</span>
5 0
3 years ago
Read 2 more answers
Ethylenediamine (en) forms an octahedral complex with Ni2+(aq) with the formula [Ni(en)3]2+. Ni2+(aq) + 3 en ⇌ [Ni(en)3]2+(aq) K
yawa3891 [41]

Answer:

3.125\times 10^{-19} mol/L is the concentration of Ni^{2+}(aq) in the solution.

Explanation:

Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)

Concentration of nickel ion = [Ni^{2+}]=x

Concentration of nickel complex= [[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L

Concentration of ethylenediamine = [en]=\frac{0.80 mol}{2 L}=0.40 mol/L

The formation constant of the complex = K_f=4.0\times 10^{18}

The expression of formation constant is given as:

K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}

4.0\times 10^{18}=\frac{0.08 mol/L}{x\times (0.40 mol/L)^3}

x=\frac{0.08 mol/L}{4.0\times 10^{18}\times (0.40 mol/L)^3}

x=3.125\times 10^{-19} mol/L

3.125\times 10^{-19} mol/L is the concentration of Ni^{2+}(aq) in the solution.

6 0
3 years ago
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