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3 years ago

Critical practices to control the amount of solid waste include __________.

Chemistry

2 answers:

Sindrei [870]3 years ago

5 0

Answer:

There are a number of critical practices that can be employed in the modern world to reduce the amount of solid waste. These include recycling materials such as paper, plastic, glass, and metals which can easily be turned into new objects, and reusing objects that do not have to be disposable for health reasons. So it's A

Explanation:

erica [24]3 years ago

3 0

Reduce, reuse, recycle, compost. I think hope its helpful.

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At equilibrium, ________. At equilibrium, ________. all chemical reactions have ceased the rate constants of the forward and rev

balandron [24]

Answer:

At equilibrium, the rate of the forward, and the reverse reactions are equal.

Explanation:

In an equilibrium chemical reaction, the rate of forward reaction, is equal to the rate of reverse reaction. Note that the reactions does not cease at equilibrium, but rather, the reactants are converted to product, at the same rate at which the product is also being converted into the reactants in the reaction. When chemical equilibrium is reached, a careful calculation of the value of equilibrium constant is approximately equal to 1.

NB: If the value of equilibrium constant is far far greater than 1, then the reaction will favors more of the forward reaction, and if far far less than 1, the reaction will favor more of the reverse reaction.

6 0

3 years ago

Solvent: Will give brainliest

Mnenie [13.5K]

Answer:

Solvent is 1.00 liter of water.

6 0

3 years ago

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Stormwater that cannot be directed back into rivers or streams is stored in

just olya [345]

Soil can get swept away into the water (by wind, gravity, rain, etc) and can increase the water level of the body of water.

Another thing is that if the soil was infected with pesticides from farmer's crops then the toxins that were in the soil can pollute the body of water. hope this helps!!

6 0

3 years ago

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During studies of the reaction below,

Leni [432]

Answer: The percent yield of the nitrogen gas is 11.53 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NO:

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

$\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol$

For $N_2O_4$ :

Given mass of $N_2O_4$ = 102.1 g

Molar mass of $N_2O_4$ = 92 g/mol

Putting values in equation 1, we get:

$\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.11mol$

For the given chemical reactions:

$2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)$ ......(2)

$N_2H_4(l)+2N_2O_4(l)\rightarrow 6NO(g)+2H_2O(g)$ .......(3)

Calculating the experimental yield of nitrogen gas:

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of $N_2O_4$

So, 0.383 moles of NO will be produced from = $\frac{2}{6}\times 0.383=0.128mol$ of $N_2O_4$

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 0.128 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 0.128=0.384mol$ of nitrogen gas

Now, calculating the experimental yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.384 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$0.384mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.384mol\times 28g/mol)=10.75g$

Calculating the theoretical yield of nitrogen gas:

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 1.11 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 1.11=3.33mol$ of nitrogen gas

Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 3.33 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$3.33mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.33mol\times 28g/mol)=93.24g$

To calculate the percentage yield of nitrogen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 93.24 g

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{10.75g}{93.24g}\times 100\\\\\% \text{yield of nitrogen gas}=11.53\%$

Hence, the percent yield of the nitrogen gas is 11.53 %.

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3 years ago

Which term refers to the region of an atom where an electron is most likely to be found?

mafiozo [28]

The term referring to that region would be "orbital".

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3 years ago

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