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poizon [28]
2 years ago
9

What two processes below are MOST SIMILAR?

Chemistry
2 answers:
mars1129 [50]2 years ago
8 0
The first one in my opinion
SpyIntel [72]2 years ago
4 0
The last one is the most similar in my opinion
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How many milliliters of a 2.00 M KBr are needed to prepare 250.0 mL of 0.60 M KBr?
PIT_PIT [208]

Answer:

75 ml

Explanation:

I did it on a Solution Calculator online.

Heres the link to the calculator to do it yourself if you want:

- https://www.sigmaaldrich.com/chemistry/stockroom-reagents/learning-center/technical-library/solution-dilution-calculator.html

7 0
3 years ago
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He gas is contained in a one-liter flask that is connected to an empty one-liter flask with a closed stopcock between the two fl
Step2247 [10]

Answer:

?H is positive and ?S is negative

Explanation:

3 0
4 years ago
What is the molarity of 20.2 g of potassium nitrate, KNO3, in enough water to make 250.0 mL of solution
KATRIN_1 [288]

Answer:

0.8M

Explanation:

CM=n/V

8 0
3 years ago
How much work must be done on a 5-kg snowboard to increase its speed from 2 m/s to 4 m/s?
Alekssandra [29.7K]

Answer:

Option C = 30 j

Explanation:

Given data:

mass of snowboard = 5 Kg

Initial speed = 2 m/s

final speed = 4 m/s

work done = ΔE= ?

ΔE= change in kinetic energy

Solution:

Formula:

K.E (initial) = 1/2 × mv²

K.E (initial)  = 1/2 × 5 Kg . (2m/s)²

K.E (initial) = 1/2 × 20 Kg.m²/s²

K.E (initial) = 10 Kg.m²/s² or 10 J      

   Kg.m²/s² = J

K.E (finial) = 1/2 × mv²

K.E (finial) = 1/2 ×  5 Kg . (4m/s)²

K.E (finial) = 1/2 ×  5 Kg . 16 m²/s²

K.E (finial) = 1/2 × 80 Kg.m²/s²

K.E (finial) = 40 Kg.m²/s²  or 40 J

work done =  ΔE = K.E (finial) - K.E (initial)

work done =  ΔE = 40 J - 10 J

work done =  ΔE = 30 J

6 0
3 years ago
What is the main idea of the article, "Great Wall of China"?
grin007 [14]

well, without the article, I'm guessing the main idea is about the Great Wall of China

7 0
3 years ago
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