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aleksandr82 [10.1K]
3 years ago
7

Write a hypothesis about the effect of temperature and surface area on the rate of chemical reactions using this format: “If . .

. then . . . because. . . .” Be sure to answer the lesson question, “How do the factors of temperature and surface area affect the rate of chemical reactions?”
Physics
2 answers:
Mandarinka [93]3 years ago
7 0
If temperature and surface area increase, then the time it takes for sodium bicarbonate to completely dissolve will decrease, because increasing both factors increases the rate of a chemical reaction.
prisoha [69]3 years ago
5 0

Answer: If temperature increases then the rate increases because energy of the particles increases.

If surface area increases then the rate increases because more contact of molecules is possible.

Explanation:

Temperature : Increasing the temperature means more reactant molecules will have sufficient energies to cross the energy barrier and thus the number of effective collisions increases, thus leading to more products and increased rate of reaction.

Surface area of the reactants: If the reactants are present in smaller size, more reactants can react as decreasing the size increases the surface area of the reactants which will enhance the contact of molecules.Hence, more products will form leading to increased rate of reaction.

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What must the driver do when approaching an intersection and seeing the traffic light turn from green to yellow?
NISA [10]
The person should start to slow down but if close enough or in the intersection go threw. Otherwise come to a complete stop until the light turns green again
4 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.33 times a second. A tack is stuck in the tire a
Troyanec [42]

Answer: 6.47m/s

Explanation:

The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.

The time taken for 1 revolution is, 1/3.33 = 0.30s

velocity of the wheel = d/t

Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t

V = 2π * 0.309 / 0.3

V = 1.94/0.3

V = 6.47m/s

The tangential speed of the tack is 6.47m/s

7 0
3 years ago
Read 2 more answers
If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun
inessss [21]

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

s=-8t^2+32t...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

\dfrac{ds}{dt}=0

\dfrac{d(-8t^2+32t)}{dt}=0

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

s=-8(2)^2+32(2)

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.

6 0
4 years ago
Light from a fluorescent lamp is observed through a cloud of cool nitrogen gas. Again, two students are having a discussion abou
Vadim26 [7]

Answer:

From the previous explanation Student No. 1 has the correct explanation

Explanation:

When the fluorescent lamp emits a light it has the shape of its emission spectrum, this light collides with the atoms of Nitrogen and excites it, so these wavelengths disappear, lacking in the spectrum seen by the observed, for which we would see an absorption spectrum

The nitrogen that was exited after a short time is given away in its emission lines, in general there are many lines, so the excitation energy is divided between the different emission lines, which must be weak

From the previous explanation Student No. 1 has the correct explanation

8 0
3 years ago
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
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