What is the overall ionic equation for zinc + silver nitrate?

Chemistry

1 answer:

3241004551 [841]3 years ago

7 0

Answer:

Zn(NO₃)₂

Explanation:

this single replacement reaction will produce silver metal, Ag , and aqueous zinc nitrate, Zn(NO3)2 . Zinc is above silver is the metal reactivity series, so it will replace silver in silver nitrate

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Use the ruler to determine the length of this object. Record your answer to the nearest tenth. The object is ____ long.

VLD [36.1K]

The object is 2.7 cm long

7 0

3 years ago

Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2

Marizza181 [45]

Answer:

$\Delta _rH=-1124.14kJ/mol$

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

$2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)$

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

$\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}$

In such a way, by using the NIST database, we find that:

$\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol$

Thus, we plug in the enthalpies of formation to obtain:

$\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol$

Best regards!

8 0

3 years ago

The hydrogen atom is not actually electronegative enough to form bonds to xenon. were the xenon-hydrogen bond to exist, what wou

Marat540 [252]

Answer : The structure of $XeH_4$ will be square-planar.

Explanation :

In the given molecule $XeH_4$ , 'Xe' is the central atom and 'H' is the terminal atom.

Xenon has 8 valence electrons and hydrogen has 1 valence electron. Therefore, the total number of valence electrons are 8 + 4(1) = 12 electrons.

The number of electrons used in Xe-H bonding = 8 electrons

The remaining electrons which are used as lone pair on central atom (Xe) = 12 - 8 = 4 electrons

There are 4 bonding pairs and 2 lone pairs of electrons, they will be arranged in the octahedral arrangement around the central atom with 2 lone pairs of electrons on central atom. The lone pairs are arranged linearly across the central atom. The resulting structure will be square-planar.

The structure of $XeH_4$ is shown below.