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3 years ago

Do all the molecules contain only single bonds?

Chemistry

2 answers:

german3 years ago

7 0

No. All molecules do not contain only single bonds.

For example, the molecules of carbon dioxide (O=C=O) and ethylene (H₂C=CH₂) contain double bonds.

The molecules of hydrogen cyanide (H-C≡N) and acetylene (H-C≡C-H) contain triple bonds.

lesya692 [45]3 years ago

3 0

No not all molecules contain only single bonds because they all have different shapes

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Tin hydrogenooxolate formula

vladimir2022 [97]

Answer:

Tin(IV) Hydrogen Oxalate. Alias: Stannic Hydrogen Oxalate. Formula: Sn(HC2O4)4. Molar Mass: 474.8178. :: Chemistry Applications:: Chemical Elements, Periodic Table.

Explanation:

3 0

3 years ago

Read 2 more answers

Convert the composition of the following alloy from atom percent to weight percent (a) 44.9 at% of silver, (b) 46.3 at% of gold,

vesna_86 [32]

Answer:

Mass percentage of gold : 33.35%

Mass percentage of silver: 62.80%

Mass percentage of copper : 3.85%

Explanation:

$N=n\times N_A$

Where : N = Number of atoms

n = moles

$N_A=6.022\times 10^{23}$ = Avogadro number

Let the total atoms present in the alloy be 100.

Atom percentage of silver = 44.9%

Atoms of of silver = 44.9% of 100 atoms = 44.9

Atomic weight of silver = 107.87 g/mol

Mass of 44.9 silver atoms = 107.87 g/mol × 44.9 = 4,843.363 g/mol

Atom percentage of gold= 46.3%

Atoms of of gold = 46.3% of 100 atoms = 46.3

Atomic weight of gold = 196.97 g/mol

Mass of 44.9 gold atoms = 196.97 g/mol × 46.3 = 9,119.711 g/mol

Atom percentage of copper = 8.8 %

Atoms of of copper = 8.8% of 100 atoms = 8.8

Atomic weight of copper = 63.55 g/mol

Mass of 44.9 copper atoms = 63.55 g/mol × 8.8 = 559.24 g/mol

Total mass of an alloy :4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol

Mass percentage of gold :

$\frac{4,843.363 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=33.35 \%$

Mass percentage of silver:

$\frac{9,119.711 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=62.80\%$

Mass percentage of copper :

$\frac{559.24 g/mol}{4,843.363 g/mol + 9,119.711 g/mol + 559.24 g/mol}\times 100=3.85\%$

7 0

3 years ago

How many grams of acetylene are produced by adding water to 5.00 g CaC2?

Murljashka [212]

Answer:

2.03125g of acetylene

Explanation:

First thing's first, we have to write out the balanced chemical equation;

CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)

Water is in excess, so CAC2 is our limiting reactant. i.e it determines the amount of product that would be formed.

1 mol of CaC2 produces 1 mol of C2H2

In terms of mass;

Mass = Number of moles * Molar mass

where the molar mass of the elements are;

Ca = 40g/mol

C = 12g/mol

H = 1g/mol

CaC2 = 40+ (2*12) = 64g/mol

C2H2 =( 2 * 12) + ( 2 * 1) = 26g/mol

64g (1 * 64g/mol) of CaC2 produces 26g ( 1mol * 26g/mol) of C2H2

5g would produce x?

64 = 26

5 = x

Upon solving for x we have;

x = (5 * 26) / 64

x = 2.03125g

5 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

Answer: The standard enthalpy change of the reaction is coming out to be -16.3 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

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

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0

3 years ago

A gas occupying a volume of 656.0 mL at a pressure of 0.884 atm is allowed to expand at constant temperature until its pressure

7nadin3 [17]

Answer:

1.14 × 10³ mL

Explanation:

Step 1: Given data

Initial volume of the gas (V₁): 656.0 mL

Initial pressure of the gas (P₁): 0.884 atm

Final volume of the gas (V₂): ?

Final pressure of the gas (P₂): 0.510 atm

Step 2: Calculate the final volume of the gas

If we assume ideal behavior, we can calculate the final volume of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 0.884 atm × 656.0 mL/0.510 atm = 1.14 × 10³ mL

5 0

3 years ago