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3 years ago

The product or products of a reaction are determined by the type of reaction taking place. true or false

Physics

1 answer:

Bess [88]3 years ago

8 0

Answer:

True

Explanation:

The products of a reaction are determined by the type of chemical reaction that are taking place. This is very true.

In chemical reactions, bonds are broken and atoms re-arranged to form new products.

By virtue of this, we can predict and know the kind of permissible combinations that can take place.
There are different kinds of chemical reaction.
They are synthesis, decomposition, single displacement, double displacement reactions e.t.c
Knowing these reactions gives insight into the likely products we can obtain.

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A wave is travelling at 3000 m/s has a wavelength of 1 m.

Levart [38]

Answer:

a] 3000hz

b]3.33 × 10⁻⁴

ci]300

ii] 3000

iii]60,000

Explanation:

3 0

2 years ago

A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit

lapo4ka [179]

Answer:

The angle is $\theta = 15.48^o$

Explanation:

From the question we are told that

The distance of the dartboard from the dart is $d = 3.66 \ m$

The time taken is $t = 0.455 \ s$

The horizontal component of the speed of the dart is mathematically represented as

$u_x = ucos \theta$

where u is the the velocity at dart is lunched

$distance = velocity \ in \ the\ x-direction * time$

substituting values

$3.66 = ucos \theta * (0.455)$

=> $ucos \theta = 8.04 \ m/s$

From projectile kinematics the time taken by the dart can be mathematically represented as

$t = \frac{2usin \theta }{g}$

=> $usin \theta = \frac{g * t}{2 }$

$usin \theta = \frac{9.8 * 0.455}{2 }$

$usin \theta = 2.23$

=> $tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}$

$\theta = tan^{-1} [0.277]$

$\theta = 15.48^o$

4 0

2 years ago

How much force is required to accelerate a 2 kg mass at 3 m/s2

swat32

Force = (mass) x (acceleration) Newton's second law of motion.

Force = (2 kg) x (3 m/s²) = 6 newtons.

3 0

3 years ago

John flies directly east for 20km, then turns to the north and flies for another 10 km before dodging a flock of geese. what’s t

KIM [24]

The distance is 30 km and the displacement is 22.4 km North East

7 0

2 years ago

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

7 0

2 years ago