The production of water proceeds according to the following equation.2H2(g) + O2(g) → 2H2O(g)

Which describes a way to speed up the collisions between hydrogen and oxygen molecules to produce more water?

a. Use a less-intense source of heat on the reactants.
b. Maintain the same temperature of the reactants.
c. Place the reactants in a smaller container.
d. Reduce the concentration of the reactants.

Answers

Answer 1
Answer: The production of water proceeds according to the following equation. 2H2(g) + O2(g) → 2H2O(g). Which describes a way to speed up the collisions between hydrogen and oxygen molecules to produce more water?

Answer: Out of all the options shown above the one that best describes a way to speed up the collisions between hydrogen and oxygen molecules to produce water is answer choice C) Place the reactants in a smaller container.

I hope it helps, Regards.
Answer 2
Answer:

Answer: C place the reactants in a smaller container

Explanation: took the quiz


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What is 850 torr= ? Atm

Answers

850 torr. times 1 atm./760 torr. or 850/760. So, 1.11842105 atm.

What mass of solute is needed to prepare each of the following solutions? a. 3.00 L of 0.125 M K2SO4 b. 550 mL of 0.015 M NaF c. 700 mL of 0.350 M C6H12O6

Answers

a.
M = 0.125 M
V = 3.0 L
M=n/V(L) ⇒ n = 0.125 mol/L *3.0L = 0.375 mol

MM of K2SO4

K2 = 2* 39 = 78 g/mol
S= 32 g/mol
O4 = 4*16 = 64 g/mol

MM = 78g/mol + 32g/mol + 64g/mol =174 g/mol

mass = n*MM = 0.375 mol*174 g/mol = 65.25 g

b.
M= 0.015 M
V = 0.55L

n = 0.015 mol/L * 0.55L = 0.00825 mol

MM NaF

Na: 23g/mol
F: 19 g/mol

MM = 23g/mol + 19g/mol = 42g/mol

mass = n * MM = 0.00825 mol * 42g/mol = 0.3465 g

c.

n = M*V(L) = 0.350 mol/L * 0,700L = 0.245 mol
MM = 6*12 g/mol + 12*1g/mol + 6*16g/mol = 180 g/mol

mass = n*MM = 0.245 mol * 180g/mol = 44.1 g
 

MATCH THE FOLLOWING!!!(match each instrument with its description)

1. weather maps/satellite
2. Doppler radar
3. radiosonde
4. wind vane
5. psychrometer
__________________________________________________________________________
A. uses two thermometers; a dry and wet bulb
B. points in the direction winds are coming from
C. uses frequency waves to track particle movement
D. used to help track air masses and fronts
E. carried with a weather balloon

Answers

Answer: The correct matching sequence is:

1 -D , 2-C, 3-E, 4-B , 5-A

Explanation:

Weather maps/satellite:These are the satellites used to monitor the climate on the Earth in day and night.Like moc=vment of clouds, El Nino. La Nina ertc

Doppler radar  is a system used to track the velocity or motion of the rain , storm or precipitation.It emit microwave signals to check the change in frequency of the wave due to the motion of the particles.

Radiosonde is an instrument carried away by balloon or any other means into different layers of atmosphere and sends information via radio.

Wind vane is an instrument used to show the direction of the wind. P

Psychrometer  is an instrument used to measure atmospheric humidity.It consist of two thermometer that is wet bulb thermometer and dry bulb thermometer.

Match each weather instrument with its description.

  1. weather maps/satellite - used to help track air masses and fronts
  2. Doppler radar - uses frequency waves to track particle movement
  3. radiosonde - carried with a weather balloon
  4. wind vane - points in the direction winds are coming from
  5. psychrometer - uses two thermometers; a dry and wet bulb

Why are fibers birefringentbecause of the chemical makeup of the fiber

because the fiber is a natural fiber

because the fiber has long range order

because the fiber has a refractive index greater than 1.52

because the refractive index is less than 1.52

Answers

Answer:

Because of the chemical makeup of the fiber.

Explanation:

The birefringence is the difference between the refractive index of a fiber in a direction at parallel to the fiber axis and in a direction at right angles. It is caused by asymmetric core shape for the two axes, and majorly one of the manufacturing imperfections. It helps in controlling polarization property inside the fiber.

Inevitable manufacturing imperfections in optical fiber leads to birefringence, which is one cause of pulse broadening in fiber-optic communications. Such imperfections can be geometrical (lack of circular symmetry), due to stress applied to the optical fiber and/or due to bending of the fiber.

A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is thelimiting reactant and how much excess reactant remains after the reaction
has stopped?

Answers

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Given data:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Excess reactant's amount left = ?

Solution:

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                       NH₃          :            N₂

                         4             :             2

                       0.12           :           2/4×0.12 = 0.06

                       NH₃         :            H₂O

                         4            :             6

                         0.12       :           6/4×0.12 = 0.18

                       

                        O₂            :            N₂

                         3             :             2

                       0.125        :           2/3×0.125 = 0.08

                         O₂           :            H₂O

                         3              :             6

                         0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                         NH₃          :             O₂

                            4            :              3

                            0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

             

Final answer:

The limiting reactant in the reaction between Ammonia and Oxygen is Ammonia (NH3). All of the Oxygen is used up in the reaction, so no excess reactant remains.

Explanation:

This question involves a concept in chemistry known as limiting reactants and stoichiometry. The balanced chemical reaction between Ammonia (NH3) and Oxygen (O2) is: 4NH3 + 5O2 -> 4NO + 6H2O. This indicates that 4 moles of NH3 react with 5 moles of O2.

To find the limiting reactant, you first need to convert the grams of your reactants to moles. The molar mass of NH3 is approximately 17.0g/mol, and the molar mass of O2 is 32.0g/mol. Therefore, you have 2.00g/17.0g/mol = 0.118 moles of NH3 and 4.00g/32.0g/mol = 0.125 moles of O2.

Since 5 moles of O2 are needed for every 4 moles of NH3, and we have slightly more O2 than NH3, the limiting reactant is NH3. To find the amount of excess reactant, we determine how much O2 actually reacted by multiplying (0.118 moles NH3)*(5 moles O2/4 moles NH3) = 0.1475 moles O2. The original amount of O2 was 0.125 moles, so the amount left over is 0.125 - 0.1475, which is a negative number and thus not possible. This confirms that O2 is the excess reactant, although it entirely reacted. Hence, no excess reactant remains.

Learn more about Limiting Reactant here:

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When 20. milliliters of 1.0 M HCL is diluted to a total volume of 60. milliliters, the concentration of the resulting solution is 1. 1.0 M 2. 0.50 M 3. 0.33 M 4. 0.25 M

Answers

The concentration of the resulting solution is 0.33 M. The correct option is 3. 0.33 M

Dilution

From the question, we are to determine the concentration of the resulting solution

From the dilution law,

M₁V₁ = M₂V₂

Where M₁ is the initial concentration

V₁ is the initial volume

M₂ is the final concentration

V₂ is the final volume

From the given information

M₁ = 1.0 M

V₁ = 20 mL

M₂ = ?

V₂ = 60 mL

Then,

1 × 20 = M₂ × 60

20 = M₂ × 60

Therefore,

M₂ = 20 ÷ 60

M₂ = 0.33 M

Hence, the concentration of the resulting solution is 0.33 M. The correct option is 3. 0.33 M

Learn more on Dilution here: brainly.com/question/5685564

C₁ * V₁ = C₂ * V₂

1.0 * 20 = C₂ * 60

20 = C₂ * 60

C₂ = 20 / 60

C₂ = 0.33 M

hope this helps!