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7nadin3 [17]
3 years ago
11

what temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution (specific heat capacity = 3

.5 J/g.c) loses 350 J of heat?
Chemistry
1 answer:
ivolga24 [154]3 years ago
3 0

Answer:

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

Explanation:

Mass of ethylene glycol = m = 100 g

Specific heat capacity of ethylene glycol = c = 3.5 J/g°C

Change in temperature of ethylene glycol = ΔT

Heat loss by the ethylene glycol = Q = 350 J

Q=mc\Delta T

\Delta T=\frac{Q}{mc}=\frac{350 J}{100 g\times 3.5 J/g^oC}

ΔT = 1°C

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

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The solubility of Z is 60 g/ 100 g water at 20 °C. How many grams of solution are produced when a saturated solution is prepared
V125BC [204]

Answer:

Saturated solution = 180 gram

Explanation:

Given:

Solubility of Z = 60 g / 100 g water

Given temperature =  20°C

Amount of water = 300 grams

Find:

Saturated solution

Computation:

Saturated solution = [Solubility of Z] × Amount of water

Saturated solution = [60 g / 100 g] × 300 grams

Saturated solution = [0.6] × 300 grams

Saturated solution = 180 gram

3 0
3 years ago
A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l)
miv72 [106K]

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)  

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

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Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V</em>

<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s)    E°red = -0.76 V</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

5 0
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How many liters of 0.85 M HCl solution would react completely with 3.5 moles Ca(OH)2?
Neporo4naja [7]
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The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i
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Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

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t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20

t=687s

7 0
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According to the bohr model of the atom, where can atoms exist? where will they ot exist
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