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r-ruslan [8.4K]
3 years ago
13

Pls HELPPP

Chemistry
2 answers:
Yuri [45]3 years ago
8 0

maybe c

Explanation:

I'm not sure but maybe

kolezko [41]3 years ago
8 0
The molecules will always move from areas of higher concentration to areas of lower concentration when you are talking about diffusion.

The solution always want to be balanced, so if you have more, let’s say salt inside the cell, then the salt will flow through the membrane to the outer less salty solution until the salt level both inside and out the cell are equal.
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Americium-241 is used in smoke detectors. it has a first order rate constant for radioactive decay of k=1.6×10−3yr−1. by contras
Helga [31]
This question is missing the part that actually asks the question. The questions that are asked are as follows:

(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.

(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.

We can use the equation for a first order rate law to find the amount of material remaining after 4 days:

[A] = [A]₀e^(-kt)

[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.

(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.

4 days x 1 year/365 days = 0.0110

A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg

The decay of americium is so slow that no noticeable change occurs over 4 days.

(b) We can simply plug in the information of iodine-125 and solve for A:

A = (1.00)e^(-0.011 x 4)
A = 0.957 mg

Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.
7 0
3 years ago
What gas is used and what is released
Shalnov [3]

Answer:

if i got it right i think it woud be a

3 0
3 years ago
How are the noble gases stable and do they exist separatly with no other element
Marrrta [24]
Yes they are stable because they follow octet rule but am not sure if they are exist separetly
7 0
3 years ago
A horizontal pipe 15 cm in diameter and 4 m long is buried in the earth at a depth of 20 cm. The pipe-wall temperature is 75 deg
Basile [38]

Explanation:

The given data for case (1) is as follows.

  h = 20 cm = 0.2 m

Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.

              Q = kA \frac{\Delta T}{L}

  where,    A = area

                  L = length

                  k = thermal conductivity = 0.8 W/m

             \Delta T = change in temperature.

Therefore, putting the given values into the above formula as follows.

            Q = kA \frac{\Delta T}{L}

                = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{0.2}

                = 168 W

For case (2), h = 180 cm = 1.8 m

Therefore, heat lost will be calculated as follows.

                       Q = kA \frac{\Delta T}{L}

                           = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{1.8}

                           = 18.67 W

Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.

8 0
3 years ago
How do you solve for the Atomic Mass in chemistry?
loris [4]
Https://www.google.com/search?q=how+to+solve+fir+atomic+mass+in+chemisty&ie=UTF-8&oe=UTF-8&hl=en-us&client=safari#kpvalbx=1
Here is the link to a great video that explains your question nicely, hope this helps.
6 0
4 years ago
Read 2 more answers
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