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3 years ago

The density of gold is 19.3 g/ml. what is gold's density in decigrams per liter?

1 answer:

4 0

To solve this question we have the density of gold expressed in gram per milliliter and we want to express in decigrams per liter. First we are going to change gram to decigram, knowing that 1 gram are 10 decigrams: 19,3 g/ml * 10 dg/1 g = 193 dg/ml Then we change milliliter to liter, knowing that 1000 ml are 1 liter: 193 dg/ml * 1000 ml/1 l = 193000 dg/l So then, density of gold is 193000 decigram/liter

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Zinc metal reacts with copper sulfate through the following

morpeh [17]

Answer:

26.9

Explanation:

5 0

2 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]

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

Explanation:

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

8 0

3 years ago

A change in velocity can occur without a change in speed * A: True B: False?

Mrrafil [7]

Answer: True

Explanation: For example, changing direction can change velocity

6 0

2 years ago

Read 2 more answers

What is the half-life of A? What will the pressure, initially 32.1 kPa, at

kodGreya [7K]

Answer:

a) 32.09 kPa

b) 32.09 kPa

Explanation:

Given data:

rate constant $= 3.56\times 10^{-7} s^{-1}$

initial pressure is = 32.1 kPa

half life of A is calculated as

$t_{1/2} = \frac{ln 2}{k}$

$t_{1/2} = \frac{ln 2}{3.56\time 10^{-7}}$

$t_{1/2} = = 1.956 \times 10^6 s$

for calculating pressure we have follwing expression

$ln p = ln P_o - kt$

$P =P_o e^{-kt}$

a) $P = 32.1 e^{-3.56\times 10^{-7} \times 10} = 32.09 kPa$

b) $P = 32.1 e^{-3.56\times 10^{-7} \times 10\times 60} = 32.09 kPa$

5 0

3 years ago

Name the following cycloalkane:

KonstantinChe [14]

B is the correct answer

8 0

2 years ago