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Stells [14]
3 years ago
7

What is true of a basic solution at room temperature? it has a ph value below 7. it has a greater concentration of hydroxide com

pared to hydronium ions. it has a distinct sour taste but an odorless gas. it can be used as a conductor in car batteries?
Chemistry
1 answer:
slega [8]3 years ago
7 0
The true statement about basic solution at room temperature is that it has a greater concentration of hydroxide compared to hydronium ions.
Basic solutions have always pH greater than 7.
Basic solutions have bitter and caustic taste.
Basic solutions are not used as conductors in car batteries, acidic electrolytes are used in car batteries.
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2 years ago
Hydrochloric acid (HCl) and potassium hydroxide (KOH) react together in an aqueous solution. Along with water, what is a product
svetoff [14.1K]

Question:

Along with water, what is a product of the reaction?

Answer:

HCl + KOH ---> KCl + H20

8 0
3 years ago
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What mass of precipitate will be produced from the reaction of 50.0 mL of 2.50 mol/L sodium hydroxide with an excess of zinc chl
amid [387]

2NaOH + ZnCl2 → 2NaCl + Zn(OH)2

Zn(OH)2 is your precipitate.

No. of moles of NaOH = 2.50X50.0÷1000 = 0.125mol

2mol of NaOH reacted to produce 1 mol of Zn(OH)2.

No. of moles of Zn(OH)2 produced when 0.125mol of NaOH reacted

= 0.125 x 1 ÷ 2

= 0.0625mol

Mass of Zn(OH)2 = 0.0625 x [65.4+2(16+1)] = 6.21g

5 0
2 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.

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3 years ago
What elements valence electrons are located in 5s25p4?<br> A. Te<br> B. Mo<br> C. Pm<br> D. Re
Katena32 [7]
Te it is correct I looked
3 0
3 years ago
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