Element (20X) in the periodic table exist in:

Humans have used rocks in the creation of tools and building materials true or flase

Anuta_ua [19.1K]

True, if you would like an example look at Indian arrow heads or early architecture all use rocks.

4 0

3 years ago

A gas at a constant volume has an initial temperature of 300.0 K and an initial pressure of 10.0 atm. What pressure does the gas

harkovskaia [24]

$\frac{500}{300} times 10=16.667 atm$

4 0

3 years ago

What attractive forces must be overcome when CsI is dissolvedin liquid HF?

bagirrra123 [75]

Answer:

The attractive forces must be overcome are :

Ion- ion interaction

Dipole - Dipole

Hydrogen Bonding

Explanation:

For the compound to dissolve the attractive forces existing between atoms of the compound must be reduced

CsI is ionic compound and its molecules are held together by ionic(electrostatic) force . These force must be weakened for its dissolution

Forces in HF :

1 .Hydrogen Bonding : In HF strong intermolecular Hydrogen Bonding exist between the electronegative F and Hydrogen

2. Dipole - dipole : HF is polar . So it is a permanent dipole and has dipole diople interaction

4 0

3 years ago

To convert 20 g of ice at -10°C to 110°C to steam you need cal of energy?

shtirl [24]

Answer:

29,200 cal = 1.22 E 5 joules

Explanation:

hope this helps

6 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