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3 years ago

Which statement best compares coastal ecosystems to open ocean ecosystems?

2 answers:

7 0

The answer is: Coastal ecosystems have more sunlight, more nutrients, and higher levels of productivity than open ocean ecosystems.

Coastal ecosystems are areas where land and water join (bays, estuaries, salt marshes, mangroves, wetlands).

In coastal ecosystems live many different types of plants and animals.

The ocean is divided into four major zones: the intertidal, neritic, oceanic and abyssal, each zone includes several ecosystems, which have adapted to specific habitats.

Anit [1.1K]3 years ago

4 0

Coastal ecosystems have more sunlight, more nutrients, and higher levels of productivity than open ocean ecosystems. These areas are shallow (lots of light) and close to river mouths (bringing high amounts of nutrients)

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Why can more water vapor (water molecules that are in the gas state of matter) be present in warm air than in cold air?

Bond [772]

Answer:

Because warm air is less dense

Explanation:

5 0

3 years ago

A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a

Semenov [28]

Answer: The metal having molar mass equal to 26.95 g/mol is Aluminium

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = $\frac{1}{2}\times 0.01670=0.00835mol$ of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol$

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

The chemical equation for the reaction of metal (forming $M^{3+}$ ion) and sulfuric acid follows:

$2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2$

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = $\frac{2}{3}\times 0.0556=0.0371mol$ of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

$0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol$

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0

3 years ago

How many mL of 0.013 M potassium hydroxide are required to reach the equivalence point in the titration of 75 mL 0.166 M hydrocy

umka21 [38]

Answer:

957.7mL

Explanation:

Using the formula below;

CaVa = CbVb

Where;

Ca = concentration of acid (M)

Va = volume of acid (mL)

Cb = concentration of base (M)

Vb = volume of base (mL)

According to the information provided in this question:

Ca = 0.166 M

Cb = 0.013 M

Va = 75mL

Vb = ?

Using CaVa = CbVb

0.166 × 75 = 0.013 × Vb

12.45 = 0.013Vb

Vb =12.45/0.013

Vb = 957.7mL

6 0

2 years ago

A balloon containing helium gas expands from 230

Anit [1.1K]

The answer for the following problem is mentioned below.

Therefore the final moles of the gas is 14.2 × $10^{-4}$ moles.

Explanation:

Given:

Initial volume ( $V_{1}$ ) = 230 ml

Final volume ( $V_{2}$ ) = 860 ml

Initial moles ( $n_{1}$ ) = 3.8 × $10^{-4}$ moles

To find:

Final moles ( $n_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

V ∝ n

$\frac{V_{1} }{V_{2} }$ = $\frac{n_{1} }{n_{2} }$

where,

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $n_{1}$ ) represents the initial moles of the gas

( $n_{2}$ ) represents the final moles of the gas

Substituting the above values;

$\frac{230}{860}$ = $\frac{3.8 * 10^-4}{n_{2} }$

$n_{2}$ = 14.2 × $10^{-4}$ moles

Therefore the final moles of the gas is 14.2 × $10^{-4}$ moles.

7 0

3 years ago

Considere un elemento "X" que posee como valencias 2, 4 y 6. En su nomenclatura tradicional, considerando la valencia cuatro, es

Ierofanga [76]

Answer:

"ite"

Explanation:

Se sabe que los elementos del grupo 16 de la tabla prioritaria muestran estados de oxidación o valencias de 2,4 y 6 respectivamente, dependiendo del compuesto formado.

Por ejemplo, el azufre forma los siguientes compuestos;

sulfato de sodio (el azufre tiene una valencia de 6)

Sulfito de sodio (el azufre tiene una valencia de 4)

Sulfuro de sodio (el azufre tiene una valencia de 2)

Por lo tanto, en compuestos en los que exhiben una valencia de 4, la terminación común es "ite"

8 0

3 years ago

Read 2 more answers