answer.
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
larisa86 [58]
1 year ago
14

Rosa was looking for patterns to help predict the products of chemical reactions. She recorded three similar decomposition react

ions in the table.
A 2-column table with 3 rows. The first column labeled reactants has entries 2 N a C l O subscript 3, 2 K C l O subscript 3, 2 L i C l O subscript 3. The second column labeled products has entries 2 N a C l + 3 O subscript 2, 3 O subscript 2 + 2 K C l, empty.

What products should she record in the last row of the table?
Physics
2 answers:
kipiarov [429]1 year ago
5 0

The products should she record in the last row of the table : <u>2LiCl + 3O₂</u>

<h3>Further explanation</h3>

There are several chemical reactions, namely:

  • Formation reaction
  • Decomposition reaction.
  • Replacement reaction.
  • Multiple replacement reactions
  • Neutralization reaction.
  • Combustion reaction.
  • Polymerization

The decomposition reaction in a chemical reaction shows the decomposition reaction of a compound into its constituent elements or compounds

Rosa recorded three similar decomposition reactions

Reactions that occur :

1. 2 NaClO₃ ⇒ 2NaCl + 3O₂

2. 2KClO₃ ⇒ 2KCl + 3O₂

3. 2LiClO₃  ⇒ 2LiCl + 3O₂

There are similarities in the decomposition that forms oxygen and chloride compounds

Learn more

homogenous mixture of two or more pure substances

brainly.com/question/1832352

Keywords : products, patterns, chemical reaction, table, decomposition reactions

#LearnWithBrainly

boyakko [2]1 year ago
4 0

Answer:

2LiCl + 3O₂

Explanation:

Hopefully this helps

You might be interested in
A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle wi
Kamila [148]

Answer:

A.)1.52cm

B.)1.18cm

Explanation:

angular speed of 120 rev/min.

cross sectional area=0.14cm²

mass=12kg

F=120±12ω²r

=120±12(120×2π/60)^2 ×0.50

=828N or 1068N

To calculate the elongation of the wire for lowest and highest point

δ=F/A

= 1068/0.5

δ=2136MPa

'E' which is the modulus of elasticity for alluminium is 70000MPa

δ=ξl=φl/E =2136×50/70000=1.52cm

δ=F/A=828/0.5

=1656MPa

δ=ξl=φl/E

=1656×50/70000=1.18cm

δ=1.18cm

6 0
2 years ago
Read 2 more answers
What type of roadway has the highest number of hazards per mile?
Oliga [24]
The roadway with the highest number of hazards is <span>city streets</span>
4 0
1 year ago
A fighter jet is catapulted off an aircraft carrier from rest to 75 m/s. If the aircraft carrier deck is 100 m long, what is the
egoroff_w [7]

The acceleration of the jet is 28.1 m/s^2

Explanation:

Since the motion of the jet is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0

v = 75 m/s

s = 100 m

Solving for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{75^2-0}{2(100)}=28.1 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0
2 years ago
Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approache
FromTheMoon [43]

Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

      W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis  

      N - W cos θ  = 0

The equation for the force of friction is

      fr = μ N

      fr = μ mg cos θ

We replace at work

     W = - μ m g cos θ  d

Mechanical energy in the lower part of the embankment

      Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

     Em_{f} = U = m g y

     W = ΔEm =  Em_{f} - Em₀

    - μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

     sin θ = y / d

     y = d sin θ

   

    - μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

     d (g syn θ + μ g cos θ) = ½ v²

     d = v²/2 g (sintea + myy cos tee)

     d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

     d = 1555.85 /7.8145

     d = 199.1 m

Let's use trigonometry to find the height

      sin 16 = y / d

      y = d sin 16

      y = 199.1 sin 16

      y = 54.9 m

8 0
2 years ago
For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




7 0
2 years ago
Read 2 more answers
Other questions:
  • Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr
    13·1 answer
  • Water is a colorless and odorless liquid. It can exist in solid, liquid, and gas states. It boils at 100 degrees C and melts at
    13·1 answer
  • The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
    12·1 answer
  • Workers do 8000 J of work on a 2000-N crate to push it up a ramp. If the ramp is 2 m high, what is the efficiency of the ramp?
    9·2 answers
  • A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision
    14·1 answer
  • Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Par
    11·1 answer
  • A charge of 5.67 x 10-18 C is placed 3.5 x 10 m away from another charge of - 3.79 x 10 "C
    6·1 answer
  • The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
    6·1 answer
  • A bicyclist is riding at a tangential speed of 13.2 m/s around a circular track. The magnitude of the centripetal force is 377 N
    14·1 answer
  • A satellite that orbits Earth with a speed of v0 must be in an orbit of radius 8RE to maintain a circular orbit, where RE is the
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!