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OverLord2011 [107]
3 years ago
7

Eutrophication is when a lake eventually become _______.

Chemistry
1 answer:
Nitella [24]3 years ago
4 0
Over flowing with BOD, which drags down the O2 levels.
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Lyle filled a glass bottle completely with water. He put the lid on tightly, then put the water in the freezer to cool it down q
svlad2 [7]

Answer:

The answer would be  It breaks them up.

Explanation:

7 0
2 years ago
Granite is an intrusive igneous rock with large crystals because it cools slowly. Where was this rock most likely formed?
Irina-Kira [14]

deep inside a volcano because it comes lava

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3 years ago
A 53. 5 mL ample of an 5. 4 % (m / v) KBr olution i diluted with water o that the final volume i 205. 0 mL Expre your anwer to t
gogolik [260]

The concentration after dilution is 1.4%.

We are aware that concentration and volume are related to each other by the formula -

C_{1} V_{1} = C_{2} V_{2}, where we have initial concentration and volume on Left Hand Side and final concentration and volume on Right Hand Side.

Keep the values to calculate final concentration.

C_{2} = (53.5 × 5.4)/205.0

Performing multiplication on Right and Side

C_{2} = 288.9/205.0

Performing division on Right Hand Side

C_{2} = 1.4%

Hence, the final concentration is 1.4%.

Learn more about concentration -

brainly.com/question/17206790

#SPJ4

The complete question is -

A 53.5 mL sample of an 5.4 % (m/v) KBr solution is diluted with water so that the final volume is 205.0 mL.

Calculate the final concentration and express your answer to two significant figures and include the appropriate units.

3 0
1 year ago
Please help me with ideas to decorate my science project board
Neporo4naja [7]
Test tubes, flasks, bunsen burners, random chemical equations

4 0
3 years ago
Read 2 more answers
How much energy (in Joules) would it take to warm 3.11 grams of gold by 7.7 oC
Reptile [31]

Answer:

It would take 3.11 J to warm 3.11 grams of gold

Explanation:

Step 1: Data given

Mass of gold = 3.11 grams

Temperature rise = 7.7 °C

Specific heat capacity of gold = 0.130 J/g°C

Step 2: Calculate the amount of energy

Q = m*c*ΔT

⇒ Q = the energy required (in Joules) = TO BE DETERMINED

⇒ m = the mass of gold = 3.11 grams

⇒ c = the specific heat of gold = 0.130 J/g°C

⇒ ΔT = The temperature rise = 7.7 °C

Q = 3.11 g * 0.130 J/g°C * 7.7 °C

Q = 3.11 J

It would take 3.11 J to warm 3.11 grams of gold

4 0
3 years ago
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