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3 years ago

Given the equation: 2Na + 2H2O → 2NaOH + H2 Which substance in this equation is a binary compound?

Chemistry

2 answers:

Blababa [14]3 years ago

8 0

By the definition of a binary compound, the answer is NaOH.

Leto [7]3 years ago

6 0

Answer: In the following equation, $H_2O$ is considered as a binary compound.

Explanation: A binary compound is considered as a compound having 2 elements only. In the following chemical equation:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

There are 2 compounds present because compound is defined as the chemical combination of two or more elements in a fixed ratio.

In the equation, compounds are: $H_2O$ and $NaOH$

$H_2O$ consists of 2 elements which are hydrogen and oxygen and hence is considered as a binary compound.

$NaOH$ consists of 3 elements which are sodium, hydrogen and oxygen and hence, cannot be considered as a binary compound.

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A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate for

laiz [17]

Answer:

CaF2 will not precipitate

Explanation:

Given

Volume of Ca(NO3)2 $= 10$ ml

Molar concentration of Ca(NO3)2 $= 0.001$

Volume of NaF $= 10$ ml

Molar concentration of NaF $= 0.0001$

Ksp for CaF2 $= 3.2 * 10^ {-11}$

CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2

Moles of calcium ion

$= 10 * 0.001\\= 0.01$

$[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}$

Moles of F- ion

$= 10 * 0.0001\\= 0.001$

$[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}$

$Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}$

Q is lesser than Ksp value of CaF2. Hence it will not precipitate

5 0

3 years ago

People mine zinc metal for use as an anticorrosive, skin protectant, and in electrical components. Zinc mining involves using ch

ankoles [38]

C I’m pretty sure. I hope this helps

3 0

3 years ago

Read 2 more answers

Which of the following compounds has the most deshielded protons?A) CH3ClB) CH3IC) CH3BrD) CH4

Korolek [52]

Given :

Some compounds :

$A)\ CH_3Cl\ B)\ CH_3I\ C)\ CH_3Br\ D)\ CH_4$ .

To Find :

Which of the following compounds has the most deshielded protons .

Solution :

Deshielded means nucleus whose chemical shift has been increased due to removal of electron density, magnetic induction, or other effects .

In simple words deshielding means the ability to shift protons .

Now , among Cl , I , Br and H . Cl is the most electron negative .

Therefore , deshielding will be more in $CH_3Cl$ .

Hence , this is the required solution .

6 0

3 years ago

Are the atoms really "sharing" electrons? Explain.

Iteru [2.4K]

Answer:

yes, in certain cases

there are different types of bondings between atoms

and in some they lend electrons to make their atom stable this type of bonding is called ionic bonding

and in covalent bond the atoms share their electrons

7 0

3 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

3 years ago