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vaieri [72.5K]
3 years ago
11

Paul made a list about alkaline earth metals.

Chemistry
2 answers:
storchak [24]3 years ago
5 0

Answer: im 99.99999% sure it is A

Explanation:

Bond [772]3 years ago
3 0

Answer:

answer is A. Alkaline metal lose two electrons not gain

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What are the flocculation basins?
Tanya [424]

Answer:

The flocculation basin often has a number of compartments with decreasing mixing speeds as the water advances through the basin. ... This compartmentalized chamber allows increasingly larger floes to form without being broken apart by the mixing blades.

7 0
3 years ago
Which of the following molecules is nonpolar?
Dvinal [7]

Answer:

CO2

Explanation:

  • There are two types of molecules
  1. Polar
  2. Non polar

Non polar molecules are insoluble in water .

7 0
3 years ago
Data for the decomposition of hydrogen peroxide at some set temperature T is provided below. The rate law depends only on the co
Viktor [21]

Answer:

0.00442 s^{-1} is the value of the rate constant.

Explanation:

2 H_2O_2\rightarrow 2 H_2O + O_2

Let the order of the reaction be x.

The rate law of the reaction can be written as:

R=k[H_2O_2]^x

1. Rate of the reaction when concentration changes from 0.882 M to 0.697 M in 0 seconds  to 60 seconds.

R=-\frac{0.697 M-0.882 M}{60 s-0 s}=0.00308 M/s

0.00308 M/s=k[0.697 M]^x..[1]

2. Rate of the reaction when concentration changes from  0.697 M to 0.566 M in 240 seconds to 360 seconds.

R=-\frac{0.236M-0.372M}{120s-60 s}=0.00227 M/s

0.00218 M/s=k[0.236 M]^x..[2]

[1] ÷ [2]

\frac{0.00308 M/s}{0.00227 M/s}=\frac{k[0.697 M]^x}{k[0.236M]^x}

Solving fro x:

x = 0.92 ≈ 1

R=k[H_2O_2]^1

0.00308 M/s=k[0.697 M]^1

k=\frac{0.00308 M/s}{[0.697 M]^1}=0.00442 s^{-1}

0.00442 s^{-1} is the value of the rate constant.

5 0
4 years ago
The charge on a sulfide ion is
Fudgin [204]

Answer: The charge on a sulfide ion is –2.

Explanation:

6 0
3 years ago
If 1.8 L of water is added to 2.5 L of a 7.0 M KOH solution, what is the molarity of the new solution?
e-lub [12.9K]
Molarity= (number of moles of that substance) / (Volume of solution in litres)

7= (number of moles of HCl) / 0.05

Thus,

Number of moles of HCl = 7 x 0.05 = 0.35

The number of moles of HCl will remain unchanged.

Now, for new solution,

0.35= 0.35 / (New volume of solution in litres)

Thus,

New volume = 1 litre

Thus, amount of water added = 1- 0.05 litre = 5.0 M

6 0
3 years ago
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