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3 years ago

In what type of environment would you most likely find Fish Species 1? Explain your answer.

Chemistry

2 answers:

KengaRu [80]3 years ago

5 0

Answer:

Fish are more likely to hold a soft bait longer. ... Signs of small fish activity can lead you to fishing hot spots. ... Cover can be loosely defined as anything that will hide or protect the fish.

Explanation:

Fishes live in virtually all aquatic habitats. Different species of fish are adapted for different habitats: rocky shores, coral reefs, kelp forests, rivers and streams, lakes and ponds, under sea ice, the deep sea, and other environments of fresh, salt, and brackish water.

VikaD [51]3 years ago

4 0

Most fishes live in aquatic environments such as the seas , rivers, and oceans

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How is burning magnesium different from burning methane?

Paha777 [63]

I believe/thought they were very similar due to the fact that they both undergo a process called “oxidation” where they release oxygen into the atmosphere.

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3 years ago

Determine whether a gas sample was composed of hydrogen or oxygen.

Nuetrik [128]

The test for this is fairly simple.
We take a glowing match or splint near the gas sample, if the glow intensifies, oxygen is present.
If a lit splint or match goes out with a popping sound, this means that hydrogen is present.

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3 years ago

Which substance can not be broken down by a chemical change?

frosja888 [35]

(4) Zirconium
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8 0

3 years ago

A student wants to make a 0.150 M aqueous solution of silver (I) nitrate but only has 11.27 g of AgNO3. What volume of the 0.150

Usimov [2.4K]

Answer:

442.3 mL

Explanation:

Remember that Molarity is a measure of concentration in Chemistry and it's defined as the number of moles of the substance divided by liters of the solution:

$M=\frac{Moles of substance X}{Volume of the solution}$

Then, you can express 11.27 g of AgNO3 as moles of AgNO3 using the molar mass of the compound:

$11.27 g AgNO_{3} *\frac{1 mole AgNO_{3}}{169.87 g AgNO_{3}} = 0.06634 moles AgNO_{3}$

Then you can solve for the volume of the solution:

$Volume of the solution=\frac{Moles of AgNO_{3}}{M} =\frac{0.06634 mol AgNO_{3}}{0.150 M} =0.4423 L = 442.3 mL$

Hope it helps!

3 0

3 years ago

Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.

BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = 240.9

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = 144

Now we can find percentage composition / percentage by mass of oxygen.

% composition = $\frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound}$ × 100

% composition = $\frac{144}{240.9}$ × 100 = 59.776%

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0

2 years ago