All categories

3 years ago

Identify the spectator ions in the following molecular equation.

1 answer:

3 0

We can rewrite the equation KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq) into its net ionic equation into
K+ + Br- + Ag + + NO3- = AgBr + K + NO3-only aqueous solutions can dissociate. Spectator ions are present in both sides, hence these are K+ and NO3-. THe rules of assigning oxidation numbers is to identify the number of valence electrons of elements and may be arbitrary depending on the charge of the molecule.

You might be interested in

46. True or false. Chemical reactions do not involve changes in the chemical bonds that join

Naddika [18.5K]

Answer:

Chemical reactions do not involve changes in the chemical bonds that join

atoms in compounds :

<u>False</u>

Explanation:

Chemical reaction are the reaction in which old bonds break and new bonds are formed . The formation of new bond result in formation of new compounds . This happen because new bond are result of linking different atoms by the bond.

For example : Water formation from Oxygen and Hydrogen is a chemical process :

$2H_{2}+O_{2}\rightarrow 2H_{2}O$

Original(old) bonds are :

H-H bond in H2 and O-O bonds in O2

In H2 = Hydrogen is joined to Hydrogen

IN O2 = Oxygen is joined to oxygen

New Bonds =

O-H bonds in water (H2O)

Oxygen is joined to hydrogen = New Bond formation

Hence,

<u>Chemical reactions do involve changes in the chemical bonds that join </u>

<u>atoms in compounds</u>

<u />

3 0

3 years ago

Aee all sets of political views correct? i need long answer

VMariaS [17]

Answer:

LOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOONGGGGGGGGGGGGGGGGGGGGGGGGGGG

Explanation:

8 0

3 years ago

This table has information about the heat of fusion and the heat of vaporization of different substances. A 3-column table with

blagie [28]

H₂S

<h3>Further explanation</h3>

Given

ΔH fusion and ΔH vaporization of different substances

Required

The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes

Solution

We can use the formula :

$\tt \Delta H=\dfrac{Q}{n}$

Q=heat/energy absorbed

n = moles

The heat absorbed : 58.16 kJ

moles = 3.11

so ΔH vaporization :

$\tt \Delta H_{vap}=\dfrac{58.16~kJ}{3.11~mol}\\\\\Delta H_{vap}=18.7~kJ/mol$

The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S

(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)

3 0

3 years ago

Read 2 more answers

Cirrus clouds form at a

sergey [27]

H > 6,000
height greater than 6,000

6 0

2 years ago

If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate

NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

4 0

3 years ago