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bulgar [2K]
3 years ago
7

HELP FAST! 35 POINTS! Why is flammability a chemical property?

Chemistry
1 answer:
Pavlova-9 [17]3 years ago
8 0
Flammmability is a chemical change because it changes the composition of the object. the product is very different from the rectant.
You might be interested in
Gold forms face-centered cubic crystals. The atomic radius of a gold atom is 144 pm. Consider the face of a unit cell with the n
kipiarov [429]

Answer:

The length of an edge of this unit cell is 407.294 pm

Explanation:

Face centered cubic structure contains 4 atoms in each unit cell and 12 coordination number, occupying about 74% volume of the total cell. Face centered cubic structure is known for efficient use of space for atom packing.

To determine the edge length, a relationship between the radius of the atom and edge length is used.

X = R√8

Where;

X is the length of an edge of this unit cell

R is the radius of the gold atom = 144 pm = 144 X 10⁻¹² m

X = 144 X 10⁻¹²√8

X = 407.294 X 10⁻¹² m

X = 407.294 pm

Therefore, the length of an edge of this unit cell is 407.294 pm

8 0
2 years ago
Mass of 0.432 moles of C8H9O4?
Fiesta28 [93]
Find one mole
8 C = 8 * 12 = 96
9 H = 1 * 9 = 9
4 O = 4 *16 = 64
Total = 169 

1 mol = 169 grams.
0.432 mol = x

1/0.432 = 169/x
x = 0.432 * 169
x = 73.0 grams
7 0
2 years ago
Read 2 more answers
1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.
solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2) The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = S=3.1\times 10^-2M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

                            S     2S

The solubility product of the lead(II) chloride = K_{sp}

K_{sp}=[Pb^{2+}][Cl^-]^2

K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}

The solubility product of the lead(II) chloride is 1.2\times 10^{-4}.

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion =1\timed 0.000010 M=0.000010 M

Solubility of aluminium hydroxide in aluminum nitrate solution = S

Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)

                            S     3S

The solubility product of the aluminium nitrate = K_{sp}=1.0\times 10^{-33}

K_{sp}=[Al^{3+}][OH^-]^3

1.0\times 10^{-33}=(0.000010+S)\times (3S)^3

S=1.6\times 10^{-10} M

The solubility of the aluminium hydroxide is 1.6\times 10^{-10} M.

3.

Molarity=\frac{Moles}{Volume (L)}

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = \frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol

Volume of the solution = 0.250 L

[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

[Cl^-]=[NaCl]=0.00024 M

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

n=0.12 M]\times 0.250 L=0.030 mol

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M

PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of lead(II) chloride = K_{sp}=1.2\times 10^{-4}

Ionic product of the lead chloride in solution :

Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}

Q_i ( no precipitation)

The given statement is false.

3 0
3 years ago
When a 3.00 g 3.00 g sample of KBr KBr is dissolved in water in a calorimeter that has a total heat capacity of 1.36 kJ ⋅ K − 1
cupoosta [38]

Answer:

Molar heat of solution of KBr is 20.0kJ/mol

Explanation:

Molar heat of solution is defined as the energy released (negative) or absorbed (Positive) per mole of solute being dissolved in solvent.

The dissolution of KBr is:

KBr → K⁺ + Br⁻

In the calorimeter, the temperature decreases 0.370K, that means the solution absorbes energy in this process. The energy is:

q = 1.36kJK⁻¹ × 0.370K

q = 0.5032kJ

Moles of KBr in 3.00g are:

3.00g × (1mol / 119g) = 0.0252moles

Thus, molar heat of solution of KBr is:

0.5032kJ / 0.0252moles = <em>20.0kJ/mol</em>

3 0
3 years ago
What is an Ore? ufhfjfnf​
Leya [2.2K]

Answer:

ore is naturally occuring solid material from which a metal or valuable mineral can be extracted

profitably.

3 0
2 years ago
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