1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
prohojiy [21]
3 years ago
14

How many atoms of hydrogen are present in 7.63 g of ammonia? please help me ??

Chemistry
1 answer:
GarryVolchara [31]3 years ago
3 0
The atoms  of  hydrogen  that are  present  in  7.63 g  of ammonia(NH3)

find  the  moles  of NH3 =mass/molar mass
 7.63 g/ 17 g/mol = 0.449  moles

since there  is  3 atoms of H  in NH3 the  moles of  H = 0.449 x 3 = 1.347 moles

by  use  of  1 mole = 6.02 x10^23  atoms
what  about  1.347  moles

= 1.347  moles/1   moles  x 6.02 x10^23 atoms = 8.11 x10^23  atoms of Hydrogen
You might be interested in
How many grams of carbon dioxide are produced from 0.98 mol of Fe3O4?
nexus9112 [7]
Hello

the answer is 43.129310000000004

Have a nice day
4 0
3 years ago
In full detail, explain what happens during the electrolysis of a NaCl brine? Be sure to identify what is being oxidized and wha
yuradex [85]

Answer:

Explanation:

Electrolysis of aqeous sodium chloride(NaCl)

Electrolysis is a process that converts electrical energy into chemical energy.

Electrolytic processes involves three major steps:

1. Ionization of electrolyte and water

2. Migration of ions to electrodes

3. Discharge of ions at the electrodes.

For the Electrolysis of brine, we follow these three steps:

1. Ionization of the aqeous brine solution:

NaCl → Na⁺ + Cl⁻

H₂O ⇄H⁺ + OH⁻

2. Migration of ions to the electrodes

The positive charges Na⁺ and H⁺ would both go to the cathode which is the negatively charged electrode

The negative charges Cl⁻ and OH⁻ migrates to the anode which are the positively charged electrodes. The anode is positively charged electrode.

3. Discharge of ions at the electrodes.

The preferential discharge of ions is based on the activity series and concentration of the ions.

On the activity series H is lower and it discharges preferentially to Na in the cathode:

2H⁺ + 2e⁻ → H₂

At this electrode, the cathode, reduction occurs and H⁺ ions are reduced.

At the anode Cl⁻ and OH⁻ migrates. But Cl⁻ is discharged preferentially due to its higher concentration.

2Cl⁻ ⇄ Cl₂ + 2e⁻

This is the oxidation half and Cl is oxidized

3 0
3 years ago
The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance b
SSSSS [86.1K]

Answer:

the stronger light 5.5 m apart from the total illumination​

Explanation:

From the problem's statement , the following equation can be deducted:

I= k/r²

where I = intensity of illumination , r= distance between the point and the light source , k = constant of proportionality

denoting 1 as the stronger light and 2 as the weaker light

I₁= k/r₁²

I₂= k/r₂²

dividing both equations

I₂/I₁ = r₁²/r₂²=(r₁/r₂)²

solving for r₁

r₁ = r₂ * √(I₂/I₁)

since we are on the line between the two light​ sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus

r₂ = r₁ + d

then

r₁ = (r₁ + d)* √(I₂/I₁)

r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)

r₁*(1-√(I₂/I₁)) =  d*√(I₂/I₁)

r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁))  =

r₁ = d/[√(I₁/I₂)-1)]

since the stronger light is 9 times more intense than the weaker

I₁= 9*I₂ → I₁/I₂ = 9 →√(I₁/I₂)= 3

then since d=11 m

r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m

r₁ = 5.5 m

therefore the stronger light 5.5 m apart from the total illumination​

5 0
2 years ago
Why is a chemical equilibrium described as dynamic? Because maximum randomness has been achieved. Because the pressure and tempe
777dan777 [17]
Because both reactants and products continue to form. Option (c) is the answer.
5 0
3 years ago
Read 2 more answers
A mixture of pure agcl and pure agbr is found to contain 60.94% ag by mass. what are the mass percents of cl and br in the mixtu
Annette [7]
First step is to calculate the mass of Ag in each compound separately:
From the periodic table: 
molar mass of Ag is 107.87 gm
molar mass of Cl is 35.45 gm
molar mass of Br is 79.9 gm
For AgCl, mass % of Ag = [107.87/143.32] x 100 = 75.26%
For AgBr, mass % of Ag = [107.87/187.77] x 100 = 57.45 %

Second step is to calculate the mass % of each compound in the mixture:
Assume mass % of AgCl is y and that of AgBr is (1-y) as the total percentage is 100% or 1
0.6094 = 0.7526 y + 0.5745 (1-y)
y = 0.8716
This means that the mixture is almost 87% AgCl and 13% AgBr
The mass % of chlorine and bromine together is (100%-60.94%) which is 39.06%
mass % of chlorine = (1-0.6094)(0.8716) x 100 = 34.044%
mass % of bromine = 39.04 - 34.044 = 5.056%
7 0
3 years ago
Other questions:
  • Write a balanced combustion reaction for C2H2 (acetylene). How did you know which products should be formed? Explain your reason
    9·1 answer
  • Which organic compound is isometric with at least one aldehyde?
    12·2 answers
  • What type of compounds dissolve easily in water?
    6·2 answers
  • A 10.00 mL sample of vinegar (an aqueous solution of acetic acid) is titrated with 0.5062 M NaOH(aq) and 16.58 mL is required to
    15·1 answer
  • Should couples wanting to start a family undergo genetic testing to minimize risk to their offspring
    15·1 answer
  • The most familiar elements are typically the most elements
    14·1 answer
  • Answer the following questions as an example of physical or chemical property. (Ive answered the first 3 already)
    9·1 answer
  • I NEED HELP PLSSS- 8th grade science
    5·2 answers
  • 6.0 mol NaOH reacts with
    9·1 answer
  • Aqueous Copper (II) nitrate reacts with aqueous potassium iodide to form Copper (II) iodide solid and potassium nitrate
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!