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Pavel [41]
2 years ago
8

Which of the following causes clouds to form? (1 point)

Chemistry
2 answers:
elena-s [515]2 years ago
6 0
The answer is B, early in the morning water covered areas (lakes,ponds,puddles,etc.) will vaporize a little bit because of the heat from the sun and it will continue all day, vapors rise towards the atmosphere and since it's a lot cooler there it will condense into a cloud which is full of tiny frozen water particles. Hope this helps <span />
Anestetic [448]2 years ago
3 0

Answer:

b

Explanation:

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Should a mortar and pestle should be used for grinding only one substance at a time
MatroZZZ [7]
The mortar and pestle is most commonly used in chemical laboratories or in the kitchen. Its key function is to grind the material into smaller pieces, usually into its powdered form. It looks like that shown in the picture. The mortar is the bowl in which the material to be pounded is placed, and the pestle does the pounding.

Now, when you ask if it can only pound one at a time, my honest answer is, it depends. Depending on the size of your mortar, you could grind materials two or three at a time. But if you are concerned with contamination, then you do it one at a time, especially if you don't want them to get mixed up.

8 0
3 years ago
Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
Nataly_w [17]
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
<span>            (M)     (M)          (M/s) </span>
<span>1         0.50    0.010      3.0×10−3 </span>
<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
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D Snack Mix because they are individual parts that can easily be seperated
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