What three consumers live in the epipelagic zone?

Which one could be considered pure substances? And why

Sergeu [11.5K]

I think it would be water but im not sure! Hope that helps!!

6 0

3 years ago

A volume of 129 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside leve

mixas84 [53]

Explanation:

As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.

So, using ideal gas equation we will calculate the number of moles as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}$

= 0.0052 mol

Also, No. of moles = $\frac{mass}{\text{molar mass}}$

0.0052 mol = $\frac{mass}{2 g/mol}$

mass = 0.0104 g

As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at $25^{o}C$

$P_{\text{water vapor}}$ = 24 mm Hg

= $\frac{24}{760}$ atm

= 0.03158 atm

Now, P = $\frac{756}{760} - 0.03158$

= 0.963 atm

Hence, n = $\frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}$

= 0.0056 mol

So, mass of $H_{2}$ = 0.0056 mol × 2

= 0.01013 g (actual yield)

Therefore, calculate the percentage yield as follows.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

= $\frac{0.01013 g}{0.0104 g} \times 100$

= 97.49%

Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.

3 0

3 years ago

A 76.0-gram piece of metal at 96.0 °C is placed in 120.0 g of water in a calorimeter at 24.5 °C. The final temperature in the ca

anygoal [31]

Answer:

S(metal) = 0.66J/g°C

Explanation:

We can find specific heat of a material, S, using the equation:

q = m*S*ΔT

Where q is change in heat, m is the mass of the substance, S specific heat and ΔT change in temperature.

The heat given by the metal is equal to the heat that water absorbs, that is:

m(Metal)*S(metal)*ΔT(Metal) = m(Water)*S(water)*ΔT(water)

Where:

m(Metal) = 76.0g

S(metal) = ?

ΔT(Metal) = 96.0°C-31.0°C = 65.0°C

m(Water) = 120.0g

S(water) = 4.184J/g°C

ΔT(water) = 31.0°C-24.5°C = 6.5°C

Replacing:

76.0g*S(metal)*65.0°C = 120.0g*4.184J/g°C*6.5°C

S(metal) = 0.66J/g°C

The law of conservation applies because the energy is not been created or destroyed. The energy that the metal gives is absorbed by the water.

3 0

3 years ago

A chemist prepares a solution of sodium chloride(NaCl) by measuring out 25.4g of sodium chloride into a 100ml volumetric flask a

labwork [276]

Answer: 4 molL-1

Explanation:

Detailed solution is shown in the image attached. The number of moles of NaCl is first obtained. Since the molarity must be in units of molL-1, the volume is divided by 1000 and the formula stated in the solution is applied and the answer is given to one significant figure.

5 0

3 years ago

Read 2 more answers

What is the volume of 1g of ice in cm3?

Amanda [17]

Answer:

What is the volume of 1g of ice in cm3?

Explanation:

3 0

3 years ago