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WINSTONCH [101]
3 years ago
15

You drop an unknown substance that weighs 8.3g into a graduated cylinder with 6ml of water. The water rises to 8ml when you drop

the object in. What is the density of this unknown object?
Chemistry
1 answer:
valkas [14]3 years ago
7 0

Answer:

<h3>The answer is 4.15 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass of object = 8.3 g

volume = final volume of water - initial volume of water

volume = 8 - 6 = 2 mL

So we have

density =  \frac{8.3}{2}  \\

We have the final answer as

<h3>4.15 g/mL</h3>

Hope this helps you

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Answer:

An atom is made of up subatomic particles called protons, neutrons and electrons. The center of an atom is called the nucleus and is where the protons and neutrons are held while electrons orbit the nucleus in orbital shells. A electron has a negative charge, a proton has a positive charge, and a neutron has no charge (neutral).

The atomic number of a atom is the total amount of the atom's protons. In a neutral atom (Not an ion), the amount of electrons is the same as the protons. Therefore, the atomic number also tells the amount of electrons in the atom.

A ion is a negatively or positively charged particle due to the giving or taking of electrons with one or more atoms (Called an ionic bond). An atom that gives away electrons becomes positively charge because that atom now has more protons than neutrons. An atom that takes an electron becomes negatively charge because that atom now has more electrons than protons.

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6 0
3 years ago
. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are
olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of S^{2-}(aq.) is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

K_{sp}\text{ of }Ag_2S=8\times 10^{-51}

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K

where,

\Delta G^o = standard Gibbs free energy = ?

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[273+25]K=298K

K = equilibrium constant or solubility product = 8\times 10^{-51}

Putting values in above equation, we get:

\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ

For the given chemical equation:

Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)

The equation used to calculate Gibbs free change is of a reaction is:  

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]

The equation for the Gibbs free energy change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]

We are given:

\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ

Putting values in above equation, we get:

285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol

Hence, the standard free energy change of formation of S^{2-}(aq.) is 92.094 kJ/mol

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