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Drupady [299]
3 years ago
5

Give the names of the ocean zones from the shore to deep ocean

Chemistry
1 answer:
BabaBlast [244]3 years ago
8 0

Answer:

the intertidal zone, neritic zone, and oceanic zone.

Explanation:

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Example: Make 100 mL 0.05 M NaOH from a 1.5 M solution.<br>0.05 * 100 = 1.5 * ?​
____ [38]

Answer:

3.33

Explanation:

M= 0.05*100/1.5= 3.33

4 0
3 years ago
The Resource Conservation and Recovery Act controls hazardous waste from its creation to its disposal. Please select the best an
o-na [289]

Answer:

True

Explanation:

Resource Conservation and Recovery Act (RCRA). 1976 of United States Environmental Protection Agency(EPA) empowers EPA to control the production, transportation, storage, treatment and disposal of hazardous waste. The RCRA act was amended in 1984 and 1986 to include Waste minimization along with appropriate disposal (not in the landfill site) and tackling of petroleum hazardous waste respectively along with other waste.

7 0
3 years ago
Read 2 more answers
Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
jeyben [28]

Answer:

0.55 atm

Explanation:

First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by

n=\frac{m}{M_m}

where

m = 1.00 g is the mass of the gas

Mm = 44.0 g/mol is the molar mass of the gas

Substituting,

n=\frac{1.00 g}{44.0 g/mol}=0.0227 mol

Now we can find the pressure of the gas by using the ideal gas law:

pV=nRT

where

p is the gas pressure

V = 1.00 L is the volume

n = 0.0227 mol is the number of moles

R = 0.082 L/(atm K mol) is the gas constant

T = 25.0 C + 273 = 298 K is the temperature of the gas

Solving the formula for p, we find

p=\frac{nRT}{V}=\frac{(0.0227 mol)(0.082 L/(atm K mol))(298 K)}{1.00 L}=0.55 atm

8 0
4 years ago
How many pairs of homologous chromosomes are found in human body cells
suter [353]
There are 22 pairs of <span>homologous chromosomes are found in human body cells.</span>
3 0
3 years ago
Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem
belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

     Fe                        4.38 g

    H₂O                      2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

Hence, the value of equilibrium constant for given equation is 1.0\times 10^{-4}

6 0
3 years ago
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