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
____ [38]
3 years ago
7

One of the solid reactants was treated in a coffee grinder before adding to

Physics
1 answer:
Nikitich [7]3 years ago
7 0

Answer:

C. Surface area

Explanation:

The rate of  chemical reaction depends on various factors such as:

  • concentration and pressure
  • nature of reactants
  • temperature
  • surface area
  • presence of catalyst, etc.

Effect of surface area of reactants: the rate of a chemical reaction can be increased by increasing the the area of contact of the reacting substances. This is especially important when one or more of the reactants are solids., because only the particles of the solid that are exposed are able to take part in the reaction at each instant of time. Therefore, the greater the surface area of the solid reactant particles the faster the reaction.

The surface area of solid reactants can be increased by grinding or pelletizing, thus allowing for a greater contact between the reacting particles,

The instance in which one of the solid reactants was treated in a coffee grinder before adding to the reaction container is one way of increasing the surface area of a reactant.

You might be interested in
PLEASE HELP In a system, when potential energy decreases, then entropy also decreases.
expeople1 [14]

False it never decreases, it increases or remain the same.


7 0
3 years ago
Read 2 more answers
Marco wants to investigate the chemical properties of sodium bicarbonate, or baking soda. He plans to carry out the tests below.
SSSSS [86.1K]

Answer:

The options are:

Test 1: Mix a teaspoon of sodium bicarbonate into a cup of warm water to see if it dissolves.

Test 2: Mix a teaspoon of sodium bicarbonate into a cup of vinegar to see if it fizzes.

Test 3: Heat a teaspoon of sodium bicarbonate in the oven to see if its melting point is less than 400 degrees Fahrenheit.

Which statement provides the best assessment of his tests?

The answer is Test 2: Mix a teaspoon of sodium bicarbonate into a cup of vinegar to see if it fizzes.

This is because it will indicate the chemical property of the sodium bicarbonate as a result of its reaction with another chemical compound, a vinegar.

Tests 1 and 3 will show a physical property of the substance as a result of the physical characteristics such as solubility and melting point .

6 0
3 years ago
Two small, identical conducting spheres repel each other with a force of 0.030 N when they are 0.65 m apart. After a conducting
solong [7]

Note that the methods applied in solving this question is the appropriate method. Check the parameters you gave in the question if you did not expect a complex number for the charges. Thanks

Answer:

q_1 = 0.00000119 + j0.00000145 C \\q_2 = 0.00000119 - j0.00000145 C

Explanation:

Note: When a conducting wire was connected between the spheres, the same charge will flow through the two spheres.

The two charges were 0.65 m apart. i.e. d = 0.65 m

Force, F = 0.030 N

The force or repulsion between the two charges can be calculated using the formula:

F = \frac{kq^2}{d^2} \\\\0.030 = \frac{9 * 10^9 * q^2}{0.65^2}\\\\q = 1.19 * 10^{-6} C

Due to the wire connected between the two spheres, q_1 = q_2 = 1.19 * 10^{-6} C

The sum of the charges on the two spheres = q_1 + q_2 = 2.38 * 10^{-6} C

Note: When the conducting wire is removed, the two spheres will no longer contain similar charges but will rather share the total charge unequally

Let charge in the first sphere = q_1

Charge in the second sphere, q₂ = 2.38 * 10^{-6} - q_1

Force, F = 0.075 N

F = \frac{k q_1 q_2}{r^2} \\\\0.075 = \frac{9*10^9 *  q_1 * (2.38*10^{-6} -q_1 )}{0.65^2}\\\\3.52 * 10^{-12} = q_1 * (2.38*10^{-6} -q_1 )\\\\3.52 * 10^{-12} = 2.38*10^{-6} q_1 - q_1^2\\\\q_1^2 - (2.38*10^{-6}) q_1 + (3.52 * 10^{-12})  = 0\\

q_1 = 0.00000119 + j0.00000145 C \\q_2 = 0.00000119 - j0.00000145 C

6 0
4 years ago
Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
If the potential in a region is given by the function V = 2 x − y 2 − cos(z), what is the y-component of the electric field at t
Bess [88]

Answer:

2y

Explanation:

Electric field in terms of Electric potential is given as:

E = dV/dr(x, y, z)

Where r(x, y, z) = position in x, y, z plane

The y component of the Electric field will be:

Ey = -dV/dy

Given that

V = 2x - y² - cos(z)

dV/dy = -2y

=> E = - (-2y)

E = 2y

8 0
4 years ago
Other questions:
  • Why do animals use echolocation? Check all that apply.
    9·2 answers
  • Object 1 has a mass of 99.2 kg. Object 2 has a mass of 42.3 kg. If the 2 object pust against each other, what will be the speed
    13·1 answer
  • A 5.44 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force
    13·1 answer
  • Similar to Hippocrates, modern scientists who study etiology believe that
    9·1 answer
  • Two girl scouts are sitting in a large canoe on a still lake while at summer camp. the canoe happens to be oriented with the fro
    5·1 answer
  • Earth’s polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes essentially nothing to the moment of inertia of
    6·1 answer
  • Verify the law of moment.​
    13·2 answers
  • The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
    14·1 answer
  • A group of students must design an experiment to determine how the amplitude of a resultant wave pulse changes when two individu
    10·1 answer
  • I NEED HELP ASAPPPP!!
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!