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3 years ago

How many protons, neutrons, and electrons are present in an atom of hafnium, Hf, with a mass number of

Chemistry

1 answer:

Soloha48 [4]3 years ago

4 0

N: 106Answer:

72 protons, 106 neutrons, 72 electrons.

Explanation:

to solve this use this simple method: APEMAN.

A= atomic number, P= Protons, E= Electrons.

M= Mass, A= atomic number, N= neutrons.

1. APE (atomic number, protons, electrons) are the same. since Hf (Hafnium) atomic number is 72, Hf has 72 protonsand ele ctrons.

2. to get Neutrons it will be mass-atomic number= neutrons.(M-A=N)

Hf mass= 178. So, 178-72= 106 neutrons.

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Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

7 0

3 years ago

Write the balanced net ionic equation for the reaction that occur in the following case. Cr2(SO4)3(aq)+(NH4)2CO3(aq)→

solmaris [256]

Cr2(SO4)3(aq) + 3(NH4)2CO3(aq) → 3(NH4)2SO4(aq) + Cr2(CO3)3(s)

Ionic: 2Cr+3 + 3SO4^-2 + 6NH4+ + 3CO3^-2 ----> 6NH4+ + 3SO4^-2 + Cr2(CO3)3 (spectator ions are NH4+, SO4^-2) 

Net Ionic: 2Cr^+3(aq) + 3CO3^-2(aq) -------> Cr2(CO3)3(s)

6 0

3 years ago

Read 2 more answers

What kinds of substances besides water can be involved in hydrogen bonding?

MrMuchimi

Kind of substance besides water:
The best example of hydrogen bonding excluding water is DNA. The two strands of polymers are connected by hydrogen bonds between the nucleotide bases.

7 0

3 years ago

Ammonia can be produced in the laboratory by heating ammonium chloride

AnnyKZ [126]

Answer:

Mass = 2.89 g

Explanation:

Given data:

Mass of NH₄Cl = 8.939 g

Mass of Ca(OH)₂ = 7.48 g

Mass of ammonia produced = ?

Solution:

2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O

Number of moles of NH₄Cl:

Number of moles = mass/molar mass

Number of moles = 8.939 g / 53.5 g/mol

Number of moles = 0.17 mol

Number of moles of Ca(OH)₂ :

Number of moles = mass/molar mass

Number of moles = 7.48 g / 74.1 g/mol

Number of moles = 0.10 mol

Now we will compare the moles of ammonia with both reactant.

NH₄Cl : NH₃

2 : 2

0.17 : 0.17

Ca(OH)₂ : NH₃

1 : 2

0.10 : 2/1×0.10 = 0.2 mol

Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.17 mol × 17 g/mol

Mass = 2.89 g

6 0

2 years ago

How many moles of solute are in 250.0 mL of a 3.00 M HCl solution?

cestrela7 [59]

0.250 L*3M=0.250 L*3mol/L= 0.750 mol

8 0

3 years ago