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otez555 [7]
3 years ago
9

If you place a 4.00 g effervescent antacid pill into 85 mL (85 g ) of water in a 90 g glass, how much will the glass of water we

igh once the antacid pill has dissolved?
Chemistry
1 answer:
nikdorinn [45]3 years ago
5 0

Answer:

The glass of water weigh once the antacid pill has dissolved will be less than the 179 grams due to evolution of carbon dioxide.

Explanation:

Mass of effervescent antacid pill = 4.00 g

Mass of water = 85 g

Mass of glass = 90 g

Total mass of the substances  together : 4.00 g + 85 g + 90 g = 179 g

Effervescent antacid pill , moment it get dissolved in water carbon dioxide gas is evolves and escape from the solutions.

This means that mass of the weight of the glass of water with antacid dissolved in it will be less than 179 grams.

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Ladybugs eat aphids on plants. Ajoy counts the number of ladybugs in one square meter of his garden.
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To estimate the number of ladybugs in the entire garden, Ajoy needs the dimension of the garden.

<h3>What is a dimension?</h3>

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Hence, to estimate the number of ladybugs in the entire garden, Ajoy needs the dimension of the garden.

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4 0
2 years ago
Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32
Law Incorporation [45]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}           ............(1)

  • <u>For a:</u>

The given chemical equation follows:

2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)

<u>Oxidation half reaction:</u>   Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-       ( × 2)

<u>Reduction half reaction:</u>   3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)

We are given:

n=2\\E^o_{cell}=+1.08V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

  • <u>For b:</u>

The given chemical equation follows:

6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)

<u>Oxidation half reaction:</u>   Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-       ( × 6)

<u>Reduction half reaction:</u>   2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)

We are given:

n=6\\E^o_{cell}=-1.29V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

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3 years ago
4.5 × 1025 atoms of nickel equal how many moles?
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There should be 4.5 moles 
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What law allows calories to be determined by heat (energy) transfer from one substance to another, but it is never destroyed?
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Answer:

matter i think

Explanation:

matter i am pretty sure

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