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3 years ago

Subject: Chemistrywhoever does this right will give the brainliest <3)

Chemistry

2 answers:

RUDIKE [14]3 years ago

8 0

Answer: This looks tough

Explanation:

Andru [333]3 years ago

4 0

Answer:

1 A

3 main types of bond are

Ionic bond ( formed due to complete transfer of electron between atoms(

Covalent bond ( formed by mutual sharing of electron)

Metalic bond ( present in the metals due to mobile electrons)

1 B bond in CaO is ionic bond formation in attached image

1 C hydrogen bond with nitrogen is covelent NH3 ammonia is formed because a bond between two non metals is expected to be covalent

More their electronegativity difference between hydrogen and nitrogen is less than 1.7 that makes it covalent

Explanation:

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Metals posses varying reduction potentials. Iron can reduce Cu+2 (aq) to copper metal. Silver can reduce Au+ (aq) to gold metal.

natta225 [31]

Answer: The true statement is iron can reduce $Au^+(aq)$ to gold metal

Explanation:

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

$A+BC\rightarrow AC+B$

Metal A is more reactive than metal B.

We are given:

Iron can reduce copper, silver can reduce gold, sodium can reduce iron and copper can reduce silver metal.

The increasing order of reactivity thus follows:

$Au$

where, sodium is most reactive and gold is least reactive

For the given options:

Option 1: Copper cannot easily reduce sodium ion to sodium metal because it is less reactive.

$Cu(s)+Na^+(aq.)\rightarrow \text{ No reaction}$

Option 2: Iron cant easily reduce gold ion to gold metal because it is more reactive.

$Fe(s)+3Au^+(aq.)\rightarrow Fe^{3+}(aq.)+3Au(s)$

Option 3: Silver cannot easily reduce iron ion to iron metal because it is less reactive.

$Ag(s)+Fe^{3+}(aq.)\rightarrow \text{ No reaction}$

Hence, the true statement is iron can reduce $Au^+(aq)$ to gold metal

7 0

3 years ago

Gamma rays are negatively charged.

asambeis [7]

Answer:

True

Explanation:

8 0

3 years ago

How old is a bone if it now .3125 of C-14 when it originally had 80.0g of C-14

Taya2010 [7]

<h3>Answer:</h3>

42960 years

<h3>Explanation:</h3>

We are given;

Remaining mass of C-14 in a bone is 0.3125 g
Original mass of C-14 on the bone is 80.0 g
Half life of C-14 is 5370 years

We are required to determine the age of the bone;

Using the formula;

Remaining mass = Original mass × 0.5^n , where n is the number of half lives.

Therefore;

0.3125 g = 80.0 g × 0.5^n

3.90625 × 10^-3 = 0.5^n

Introducing logarithm on both sides;

log 3.90625 × 10^-3 = n log 0.5

Solving for n

n = log 3.90625 × 10^-3 ÷ log 0.5

= 8

Therefore, the number of half lives is 8
But, 1 half life is 5370 years
Therefore;

Age of the rock = 5370 years × 8

= 42960 years

Thus, the bone is 42960 years old

7 0

3 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

3 years ago

Substance/Specific heat capacity

nikdorinn [45]

I believe it is aluminum

8 0

3 years ago

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