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Sav [38]
2 years ago
6

for a neutralization reaction, would you expect the magnitude of q to increase, decrease, or stay the same if the concentration

of only the acid were doubled? why?
Chemistry
1 answer:
ikadub [295]2 years ago
4 0

For a neutralization reaction, the value of q(heat of neutralization) is doubled when the concentration of only the acid is doubled.

A neutralization reaction is a reaction in which an acid reacts with a base to yield salt and water. Ionically, a neutralization reaction goes as follows; H^+(aq) + OH^-(aq) ------> H20(l).

The heat of neutralization (Q) of the system depends on the concentration of the solutions. Since Q is dependent on concentration, if the concentration of any of the reactants is doubled, more heat is evolved hence Q is doubled.

Learn more: brainly.com/question/10323185

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How many particles are present in 8.0 moles of silver?
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Answer:the answer is b

Explanation:I took the test and got it right

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3 years ago
A pH scale reading 13 indicates a ______
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A pH scale reading 13 indicates a strong base.

Explanation:

From my understanding:

1 -4 is a strong acid

4 - 7 is weak acid

7 - 9 is a weak base

9 - 14 is a strong base

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2 years ago
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Place these elements in order of LOWEST electronegativity to HIGHEST.
mrs_skeptik [129]

Answer:

Barium

Molebdenum

Cobalt

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Explanation:

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8 0
3 years ago
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Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which
Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride =  0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be:= \dfrac{5 \ kg }{1590 \ kg/m^3}

= 0.0031 m³

Suppose \beta is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT

In ( \dfrac{V_2}{V_1})  = \beta (T_2-T_1)

V_2 = V_1 \times exp (\beta (T_2-T_1))

V_2 = 0.0031 \ m^3  \times exp  (1.2 \times 10^{-3} \times 20)

V_2 = 0.003175 \ m^3

However; the workdone = -PdV

W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)

W = - 7.6 J

The heat energy Q = Δ h

Q = mC_p(T_2-T_1)

Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

3 0
3 years ago
Which member of each pair is more soluble in diethyl ether? Why?<br> (c) MgBr₂(s) or CH₃CH₂MgBr(s)
White raven [17]

CH3CH2MgBr is more soluble in diethyl ether .

We know that polar solvent dissolve in polar solvent very perfectly . as diethyl ether is a polar solvent so it have dipole -dipole interaction .

Hence the compound with similar interaction can dissolve in diethyl  ether .

Here , MgBr2  is an ionic compound . there is ion-ion interactions occurs which is not similar to dipole -dipole interaction in diethyl ether .hence the solubility of MgBr2 in diethyl ether is less .

but in case of CH3CH2MgBr there are both polar and nonpolar end .CH3CH2 is the nonpolar end and MgBr is the polar end .

thus with the nonpolar end solute interact using depression forces and with polar end solute interact using dipole-dipole interaction . so CH3CH2MgBr is more soluble .

Learn more about polar solvent here :

brainly.com/question/3184550

#SPJ4

4 0
2 years ago
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