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2 years ago

How does the use of synthetic products impact the natural resource from which they are derived?

Chemistry

1 answer:

oksian1 [2.3K]2 years ago

5 0

Answer:

The series of chemical reactions that are used to change natural resources into synthetic products is called chemical synthesis. To make a "natural" product, the natural resource is not chemically changed as much. ... For example, the synthetic material plastic is made from petroleum, which is pumped out of the earth.

Explanation:

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How much heat is absorbed when 447 g of water , Cp = 4.184 J/goC, goes from 25.0 oC to 35.0 oC?

Digiron [165]

Answer:

The answer to your question is Q = 18702.5 J

Explanation:

Data

mass of water = m = 447 g

Cp = 4.184 J/g°C

Temperature 1 = T1 = 25°C

Temperature 2 = T2 = 35°C

Heat = Q = ? Joules

Process

1.- Write the formula to calculate heat

Q = mCp(T2 - T1)

2.- Substitution

Q = (447)(4.184)(35 - 25)

3.- Simplification

Q = (447)(4.184)(10)

4.- Result

Q = 18702.5 J

8 0

3 years ago

A lake in rural maine drops in ph from 7 to 5 as a result of acid rain falling into it. how has the h+ ion concentration change

Jlenok [28]

The H+ concentration in the lake has increased as a result of the acid rain.
The original pH of the lake was 7, which mean the water is neutral, but due to the acid rain, it drops to 5. This means, that the water has become acidic and how have more hydrogen ion. The H+ concentration in the lake has 100 times compares with its original pH.

3 0

3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

2 years ago

3. What is the mass of HI used to create 2.50L of a 0.48 M solution of hydroiodic acid?

Daniel [21]

Answer:

Mass = 153.48 g

Explanation:

Given data:

Volume of solution = 2.50 L

Molarity = 0.48 M

Mass required = ?

Solution:

Molarity = number of moles / volume in litter

Number of moles = Molarity × volume in litter

Number of moles = 0.48 M × 2.50 L

Number of moles = 1.2 mol

Mass of HI:

Number of moles = mass/molar mass

Mass = Number of moles × molar mass

Mass = 1.2 mol × 127.9 g/mol

Mass = 153.48 g

5 0

3 years ago

What NaCl NaCl concentration results when 249 mL 249 mL of a 0.850 M 0.850 M NaCl NaCl solution is mixed with 667 mL 667 mL of a

tamaranim1 [39]

Answer:

The answer to your question is 0.54M

Explanation:

Data

Final concentration = ?

Concentration 1 = 0.850 M

Volume 1 = 249 ml = 0.249 l

Concentration 2 = 0.420 M

Volume 2 = 0.667 M

Process

1.- Calculate the number of moles in both solutions

Number of moles 1 = Molarity 1 x Volume 1

= 0.850 x 0.249

= 0.212

Number of moles 2 = Molarity 2 x Volume 2

= 0.420 x 0.667

= 0.280

Total number of moles = 0.212 + 0.280

= 0.492

2.-Calculate the final volume

Final volume = Volume 1 + Volume 2

Final volume = 0.249 + 0.667

= 0.916 l

3.- Calculate Molarity

Molarity = 0.492 / 0.916

Molarity = 0.54

4 0

2 years ago