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2 years ago

What is one way that fish interact with a coral reef

Chemistry

2 answers:

Digiron [165]2 years ago

8 0

Answer:

Fishes play a vital role in the reef's food web, acting as both predators and prey. Their leftover food scraps and wastes provide food or nutrients for other reef inhabitants. Some species of sharks, skates, and rays live on or near the reef. ... These fish are herbivores and eat the algae within the coral.

Explanation:

LenaWriter [7]2 years ago

3 0

Answer:

The second answer. Fish hide within the coral. Hope this helps!!!!

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There are 235.5 grams of Calcium chloride dissolved in 2.5 liters of solution. What is the molarity of

Fed [463]

M= moles/L covert 235.5 grams of CaCl2 using the molar mass of the compound obtaining the number of moles. Then divide the number by 2.5 L

7 0

2 years ago

20.0 g of Nitrogen is produced when oxygen gas reacts with NO gas. 29.8 L of oxygen is required at STP to produce this 20.0 g of

Rainbow [258]

Answer:

There will be produced 30.64 grams of nitrogen dioxide

The statement is false

Explanation:

Step 1: Data given

MAss of nitrogen produced = 20.0 grams

Molar mass of N2 = 28.0 g/mol

Mass of NO = 20.0 grams

Molar mass of NO = 30.01 g/mol

Volume of O2 at STP = 29.8 L

Step 2: the balanced equation

2NO + O2 → 2N02

Step 3: Calculate moles NO

Moles NO = mass NO / molar mas NO

Moles NO = 20.0 grams / 30.01 g/mol

Moles NO = 0.666 moles

Step 4: Calculate moles O2

1 mol O2 at STP = 22.4 L

29.8 L = 29.8/22.4 = 1.33 moles

Step 5: Calculate the limiting reactant

For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

NO is the limiting reactant. IT will completely be consumed (0.666 moles).

O2 is in excess. There will react 0.666/2 = 0.333 moles O2

There will remain 1.333 - 0.333 = 0.997 moles O2

Step 6: Calculate moles NO2

For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

For 0.666 moles NO we'll have 0.666 moles NO2

Step 7: Calculate mass NO2

Mass NO2 = moles NO2 * molar mass NO2

Mass NO2 = 0.666 moles * 46.0 g/mol

Mass NO2 = 30.64 grams

There will be produced 30.64 grams of nitrogen dioxide

The statement is false

3 0

3 years ago

A buffer is made that is .10M propanoic acid (pKa = 4.9) and .05M sodium propanoate. Calculate the pH.

inna [77]

PKa= 4.9 therefore ka= 10^-4.9= 1.259x10^-5

$ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]}$

$[CH3CH2COO^-]
$ = 0.05
$[CH3CH2COOH]$ = 0.10

Therefore 1.259x10^-5 = $\frac{[H^+][0.05]}{[0.1]}$
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore $[H^+] = \frac{(1.259*10^-5)(0.1)}{0.05}$

Therefore $[H^+]= 2.513*10^-5$

pH= -log [ $H^+$ ] = -log(2.513*10^-5)= 4.59.