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uranmaximum [27]
4 years ago
13

Experimental data have shown that the rate law for the reaction 2 hgcl2(aq) + c2o4 2 -(aq) → 2 cl-(aq) + 2 co2 (g) + hg2cl2 (s)

is: rate = k[hgcl2][c2o4 2 -]2 how will the rate of reaction change if the concentration of c2o4 2 - is tripled and the concentration of hgcl2 is doubled?
Chemistry
1 answer:
NNADVOKAT [17]4 years ago
6 0
When the rate = K [ HgCl2] [C2O4 2-]^2
if the concentration of [ C2O4 2- ] is tripled so we will substitute by [3 (C2O4 2-)] instead of [ C2O4 2- ] 
and if the concentration of [HgCl2] is doubled so we will substitute by [2( HgCl2)] instead of [HgCl2] 
So the new rate will be = K [2(Hgcl2)] [3(C2O4 2-)]^2
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in order to find the molar mass of an unknown compound, a research scientist prepared a solution of 0.930 g of an unknown in 125
PtichkaEL [24]

Answer:

Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

Explanation:

Let's apply the formula for freezing point depression:

ΔT = Kf . m

ΔT = 74.2°C - 73.4°C → 0.8°C

Difference between the freezing T° of pure solvent and freezing T° of solution

Kf = Cryoscopic constant → 5.5°C/m

So, if we replace in the formula

ΔT = Kf . m → ΔT / Kf = m

0.8°C / 5.5 m/°C = m → 0.0516 mol/kg

These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg

0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:

Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol

3 0
3 years ago
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
Sugar is stirred into water tp form a solution what names to we give to the sugar and water
Nina [5.8K]
Your answer would be, Sugar is the solute, and water is the solvent. a solution is the mixture of two, or more substances.




Hope that helps!!!
7 0
4 years ago
The atomic symbols of nitrogen and oxygen are N and O, respectively. The chemical formula of a particular substance is NO2. Whic
muminat
A compound made out of 1 nitrogen atom and 2 oxygen atoms
7 0
3 years ago
Read 2 more answers
A 250-mL aqueous solution contains 1.56 mc025-1.jpg 10–5 g of methanol and has a density of 1.03 g/mL. What is the concentration
Elis [28]
The right answer for the question that is being asked and shown above is that: "B 1.5 X 1-^-5 ppm." A 250-mL aqueous solution contains 1.56 mc025-1.jpg 10–5 g of methanol and has a density of 1.03 g/mL. The concentration in ppm is that <span>1.5 X 1-^-5 ppm</span>
4 0
4 years ago
Read 2 more answers
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