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skelet666 [1.2K]
2 years ago
7

What are three ways invasive species can be introduced into an environment?

Chemistry
1 answer:
Naily [24]2 years ago
6 0

Answer:

A

B

D

(:

Explanation:

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I need help with this answer
frez [133]
A
not every expirement has control
7 0
3 years ago
How would you prepare 3.5 L of a 0.9M solution of KCl?
PolarNik [594]
V=3,5L\\
Cm=0,9M\\
M_{KCl}=74\frac{g}{mol}\\\\
C_{m}=\frac{n}{V}\\\\
n=\frac{m}{M}\\\\
C_{m}=\frac{m}{MV} \ \ \ \Rightarrow  \ \ \ m=C_{m}MV\\\\
m=0,9\grac{mol}{L}*74\frac{g}{mol}*3,5L=233,1g

B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark. 
4 0
3 years ago
Can someone please definition of atomic radius using , nucleus, valence electrons and energy.❤️
OLEGan [10]

Answer:

Explanation:

-The Atomic Radius of an element is the distance between the center of an atom

-nucleus and its outermost, or valence electrons. ... These changes are caused by the interaction between the positive charge of the protons

- nucleus and the negative charge of all the atom's electrons.

5 0
3 years ago
1 points)<br> You have 300 grams of Al2(CO3)3. How many moles are produced?
Pavlova-9 [17]

Answer:

0.5133805136 moles.

Explanation:

1 gram of Al2(Co3)3 equals 0.0017112683785004 moles, we need the amount of moles produced in 300 grams of Al2(CO3)3, so we have to multiply 1 gram of Al2(CO3)3 times 300: 0.0017112683785004 x 300, in conclusion,

300 grams of Al2(Co3)3 equals 0.5133805136.

5 0
2 years ago
RATE LAW QUESTION !
vivado [14]
In general, we have this rate law express.:

\mathrm{Rate} = k \cdot [A]^x [B]^y
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4)  by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation \Delta \mathrm{Rate} = \Delta [B]^y

3.09 = (3)^y \implies y \approx 1

To this point, \mathrm{Rate} = k \cdot [A]^x [B]^1

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for \Delta \mathrm{Rate} = \Delta [A]^x

4=  (2)^x \implies x = 2

the rate law is

Rate = k·[A]²[B]
6 0
3 years ago
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