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madreJ [45]
3 years ago
15

1.) C6H12O6 (aq) is a(an)?

Chemistry
1 answer:
Marina86 [1]3 years ago
8 0

Answer:

1) D. Dissolved Molecule

2) C. 11.30 M

Explanation:

<u><em>Q1:</em></u>

  • Glucose (C₆H₁₂O₆) is a non-polar molecule.
  • When it is dissolved in water, it will be a dissolved molecule.
  • It can not be ionized in water.
  • So, the right choice is: <em>D. Dissolved molecule.</em>

<u><em>Q2</em></u>:

  • The molarity is the no. of moles of dissolved solute in a 1.0 L of the solution.

<em>M = (n)solute (1000/V (mL) of the solution)</em>

n of LiCl = 2.60 mol, V = 230.0 mL.

∴ M = (n)solute (1000/V (mL) of the solution) = (2.60 mol)(1000)/(230.0 mL) = 11.30 M.

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The enthalpy of reaction changes somewhat with temperature. Suppose we wish to calculate ΔH for a reaction at a temperature T th
Contact [7]

Answer:

-99.8 kJ

Explanation:

We are given the methodology to answer this question, which is basically  Kirchhoff law . We just need to find the heats of formation for the reactants and products and perform the calculations.

The standard heat of reaction is

ΔrHº = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where ν are the stoichiometric coefficients in the balanced equation, and ΔfHº are the heats of formation at their  standard states.

  Compound                 ΔfHº (kJmol⁻¹)

        SO₂                             -296.8

         O₂                                    0

         SO₃                            -395.8

The balanced chemical equation is

SO₂(g) + ½O₂(g) → SO₃(g)

Thus

Δr, 298K Hº( kJmol⁻¹ ) =  1 x (-395.8) - 1 x (-296.8) = -99.0 kJmol⁻¹

Now the heat capacity of reaction  will be be given in a similar fashion:

Cp rxn = ∑ ν x Cp of products - ∑ ν x Cp of reactants

where ν is as above the stoichiometric coefficient in the balanced chemical equation.

Cprxn ( JK⁻¹mol⁻¹) = 50.7 - ( 39.9 + 1/2 x 29.4 ) = - 3.90

                         = -3.90 JK⁻¹mol⁻¹

Finally Δr,500 K Hº = Δr, 298K Hº +  CprxnΔT

Δr,500 K Hº = - 99 x 10³ J + (-3.90) JK⁻¹ ( 500 - 298 ) K = -99,787.8

                     = -99,787.8 J x 1 kJ/1000 J  = -99.8 kJ

Notice thie difference is relatively small that is why in some problems it is o.k to assume the change in enthalpy is constant over a temperature range, especially if it is a small range of temperatures.

3 0
3 years ago
O
viva [34]

Answer:

The second and third option are isotopes of Bromine

Z=  35 A = 79

A = 79  N= 44

Explanation:

Step 1: What are isotopes ?

⇒ Elements  with same atomic number (Z) ( this means the same number of electrons and protons) but a different number of neutrons (N)

The atomic number (visible on the periodic table) is the number of protons.

The atomic mass is  the sum of the protons (Z) and neutrons (N), and is showed as 'A'. So A = Z+N

If we look at the periodic table, we can see that the atomic number of bromine (Z) = 35. This means the amount of protons = 35. Since isotopes have the same amount of protons, all isotopes of bromine, have 35 protons.

1) Z = 79, A = 196

Z = protons = electrons .So this element has 79 protons, as well as 79 electrons.

196 = 79 + N ⇒ N = 196 - 79 = 117 neutrons

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Z = protons = electrons .So this element has 35 protons, as well as 35 electrons.

79 =35 + N ⇒ N = 79 - 35= 44 neutrons

⇒Since it does have 35 protons, it is an isotope of Bromine. This isotope has 44 neutrons

⇒ This is 79Br, which is a stable isotope of Bromine.

3) A=79, N = 44

Z = A - N ⇒ Z = 79 - 44 = 35

Z = protons = electrons .So this element has 35 protons, as well as 35 electrons. It also has 44 neutrons.

⇒Since it does have 35 protons, it is an isotope of Bromine, with 44 neutrons: 79Br

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Z = protons = electrons .So this element has 44 protons, as well as 44 electrons.

A = 44 + 44 ⇒ A = 88  

⇒Since it doesn't have 35 protons, it isn't an isotope of Bromine, but of Ruthenium (Ru) : 88Ru

⇔So the <u>second</u> and<u> third</u> option are isotopes of Bromine

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