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Fudgin [204]
3 years ago
5

How do i know if a reaction is exo/endothermic?

Chemistry
1 answer:
oee [108]3 years ago
8 0

Answer:

You would if a reaction is exo/endothermic because an endothermic reaction soaks up heat. An exothermic reaction releases heat.

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How many half-lives will pass by the time 1.56% of I-131 is present? B. Approximately how many days does that equal? *
serg [7]

Answer: Hmmmmm that's crazy....

There are a couple of equations one could use for this type of problem, but I find the following to be the easiest to use and to understand.

Fraction remaining (FR) = 0.5n

n = number of half lives that have elapsed

In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).

0.0156 = 0.5n

log 0.0156 = n log 0.5

-1.81 = -0.301 n

n = 6.0 half lives have elapsed

Explanation:

Just wanted to help. Hopefully it's correct wouldn't want to waster your time ;)

6 0
2 years ago
Ultraviolet radiation has a frequency of 6.8×1015 Hz. What is the energy associated with this frequency?
stellarik [79]

Use Planck's equation (E=hv) to solve.  where <span>frequency (v) of ultrviolet radiation is 6.8 × 1015 1/s.  </span><span> </span>The variable h is a constant equal to 6.63 × 10-34 J·s
E= <span>(6.8 × 1015 1/s)x(</span>6.63 × 10-34 J·s)


5 0
3 years ago
What is humidity? Please answer as quickly as possible.
Katen [24]
"a quantity representing the amount of water vapor in the atmosphere or a gas."
7 0
3 years ago
Read 2 more answers
A particle with a charge of 9.40 nC is in a uniform electric field directed to the left. Another force, in addition to the elect
juin [17]

Answer:

a. Work done by the electric force = -2.85 * ×10⁻⁵ J

b. The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. The magnitude of the electric field is 33.7kV/m

Explanation:

Given.

Charge = Q = 9.40 nC

Distance = d = 9.00 cm = 0.09m

Amount of work = 7.10×10⁻⁵ J

Kinetic energy = K = 4.25×10⁻⁵ J

a. What work was done by the electric force?

This is calculated by; change in Kinetic Energy i.e. ∆KE

∆KE = ∆K2 - ∆Kæ

Where K2 = 4.25×10⁻⁵ J

The body is released at rest, so the initial velocity is 0.

So, K1 = 0

Also, total work done = W1 + W2

Where W2 = 7.10×10⁻⁵J

So, W1 + W2 = W = K2

W1 + 7.10×10⁻⁵ = 4.25×10⁻⁵

W1 = 4.25×10⁻⁵ - 7.10×10⁻⁵

W = -2.85 * ×10⁻⁵ J

Work done by the electric force = -2.85 * ×10⁻⁵ J

b. What is the potential of the starting point with respect to the end point?

The change in potential energy is given as

W = ∆U

W = Q|V2 - V1| where V1 = 0 because the body starts from rest

So, W = QV2

Make V the Subject of the formula

V2 = W/Q

V2 = -2.85 * ×10⁻⁵ J / 9.40 nC

V2 = -2.85 * ×10⁻⁵ J / 9.40 * 10^-9C

V2 = −3031.9148936170212765957V

V2 = -3.03 * 10³ V

The potential of the starting point with respect to the end point = -3.03 * 10³ V

c. What is the magnitude of the electric field?

The magnitude of the electric field is calculated as follows;

W = -Fd = -QEd

And E = V/d

E = -3.03 * 10³ V / 0.09 m

E = −33687.943262411347517730 V/m

E = -33.7kV/m

The magnitude of the electric field is 33.7kV/m

4 0
3 years ago
Common thermoplastic polymers such as polystyrene and PVC can be _______ as often as they are reheated. A) hardened B) shortened
tangare [24]
The answer is A) hardened
5 0
2 years ago
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