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Leya [2.2K]
4 years ago
7

Instead of just clear cutting land and moving on, a local lumber business decides to start farming, planting the fields they hav

e just cut with young trees. The impact this strategy will have on the environment in sustaining the soil is best described as what?
Chemistry
1 answer:
Dafna1 [17]4 years ago
6 0

Answer:

C) Positive, since it is a sustainable plan to use a renewable resource.

Explanation:

The local lumber business, with this move, has done good for both, to the business itself, and to the environment as well. Since the amount of trees that will be cut, is immediately replaced with saplings that are taken care of, the lumber business is creating a sustainable business for the future, and also it is creating a renewable resource. The environment will also benefit from this, because the lumber business will only use certain amount of wooded area because it will have new wood constantly on the same area it is using, thus they will not have the need to spread out on a larger area and destroy the environment.

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Mrs. Borunda has a sample of oxygen gas that occupies a volume of 600 L at 400 atm pressure. What will the pressure be if she in
kotykmax [81]

Answer:

P₂ = 300 atm

Explanation:

Given that,

Initial volume, V₁ = 600 L

Initial pressure, P₁ = 400 atm

We need to find the pressure if the volume is 800 L.

We know that the relation between pressure and volume is given by :

P\propto \dfrac{1}{V}\\\\\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\\P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{600\times 400}{800}\\\\P_2=300\ atm

So, the new pressure is equal to 300 atm.

5 0
3 years ago
Please what is the name of this compound?​
wariber [46]

2,3 diethylpentene all one word. Brainliest would be much appreciated

3 0
3 years ago
Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result
Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0
3 years ago
What is a metal and a non metal​
Mrac [35]

Answer:

I assume you mean as in elements

A metal "A metal is a material that, when freshly prepared, polished, or fractured, shows a lustrous appearance, and conducts electricity and heat relatively well. Metals are typically malleable or ductile." (wiki)

a non-metal "In chemistry, a nonmetal is a chemical element that mostly lacks the characteristics of a metal. Physically, a nonmetal tends to have a relatively low melting point, boiling point, and density. A nonmetal is typically brittle when solid and usually has poor thermal conductivity and electrical conductivity." (wiki)

Explanation:

5 0
3 years ago
Read 2 more answers
At 20°C, a 0.756 M aqueous solution of ammonium chloride has a density of 1.0107 g/mL. What is the mass % of ammonium chloride i
Molodets [167]

Answer:

\%m/m=4\%

Explanation:

Hello,

In this case, by knowing that the molarity is measured in molal units which are mole per liter of solution and the by-mass percentage demands us to compute the mass of the solution, we proceed by assuming 1 L of solution:

m_{solution}=1L*\frac{1000mL}{1L}*\frac{1.0107g}{1mL} =1010.7g

Then, for 1 L of solution, we have 0.756 moles of solute (ammonium chloride), so we compute the grams for those moles by using its molar mass of 53.491 g/mol as shown below:

m_{solute}=0.756mol*\frac{53.491g}{1mol}=40.4g

Finally, we compute the by-mass percentage as shown below:

\%m/m=\frac{m_{solute}}{m_{solution}} *100\%=\frac{40.4g}{1010.7g}*100 \%\\\\\%m/m=4\%

Best regards.

6 0
3 years ago
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