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taurus [48]
2 years ago
6

A sample of a substance burns more rapidly in pure oxygen than in air. which factor is most responsible for this high rate of re

action?
Physics
1 answer:
zmey [24]2 years ago
3 0
I had to look for the options and here is my answer. Based on the given substance above that burns faster in pure oxygen than in air, the factor that is most responsible for this kind of reaction is the concentration of the reactants. This answer is based on the actual options attached to this question.
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Help please! This question is driving me crazy
Varvara68 [4.7K]

Answer:

-10.8°, or 10.8° below the +x axis

Explanation:

The x component of the resultant vector is:

x = 3.14 cos(30.0°) + 2.71 cos(-60.0°)

x = 4.07

The y component of the resultant vector is:

y = 3.14 sin(30.0°) + 2.71 sin(-60.0°)

y = -0.777

Therefore, the angle between the resultant vector and the +x axis is:

θ = atan(y / x)

θ = atan(-0.777 / 4.07)

θ = -10.8°

The angle is -10.8°, or 10.8° below the +x axis.

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3 years ago
The following is current scientific evidence supporting the nebular theory on the formation of the solar system.
Feliz [49]
A.the composition of the inner and outer planets, current observations of star formation, and the motion of the solar system I hope this helps
4 0
3 years ago
Read 2 more answers
Given the equation E = P/N , solve for P
Mars2501 [29]

Answer:

E=P/N

multiply both sides by N

P=EN

6 0
2 years ago
When monochromatic light illuminates a grating with 7000 lines per centimeter, its second order maximum is at 62.4°. what is the
zhannawk [14.2K]

When red light illuminates a grating with 7000 lines per centimeter, its second maximum is at 62.4°. What is the wavelength of this light?

ans: 633nm

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3 years ago
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If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int
vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system on the surrounding


In this problem, the work done by the system is

W=-225 kcal=-941.4 kJ

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

Q=-5 \cdot 10^2 kJ=-500 kJ

with a negative sign as well because it is released by the system.


Therefore, by using the initial equation, we find

\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ

8 0
3 years ago
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