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2 years ago

A conservation easement would be best suited to offset which of the following threats to boidiversity?

Chemistry

2 answers:

Tanzania [10]2 years ago

7 0

Habitat loss

Just took the test and overharvesting is incorrect.

Leto [7]2 years ago

3 0

Overharvesting that would be your answer!

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Balance the following equation. C3H8(g) + O2(g) → CO2(g) + H2O(l)

Arte-miy333 [17]

Explanation:

C3H8+5O2----->3CO2+4H2O

8 0

3 years ago

Is it possible to change a scientific theory?

Crazy boy [7]

Answer:

I think the first option is the answer

3 0

2 years ago

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A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 385.mg of oxalic acid H2

Vlada [557]

Answer:

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

Step 1: Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

volume = 250 mL = 0.250 L

Volume of NaOH = 123.7 mL = 0.1237 L

Molar mass H2C2O4 = 90.03 g/mol

Step 2: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

Step 3: Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

Moles H2C2O4 = 0.00428 moles

Step 4: Calculate molarity of H2C2O4

Molarity H2C2O4 = moles / volume

Molarity H2C2O4 = 0.00428 moles / 0.250 L

Molarity H2C2O4 = 0.01712 M

Step 5: Calculate molarity of NaOH

2*Ca*Va = n*Cb*Vb

⇒ with Ca = Molarity of H2C2O4 = 0.01712 M

⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M

4 0

3 years ago

Soil particles tend to be __________ charged and attract __________ charged ions.

aliina [53]

Answer:it’s abc it’s just science you know

Explanation:

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3 years ago

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vess

Tpy6a [65]

Answer:

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

Explanation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The relation of $K_c\& K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure.=98.1

$K_c$ = Equilibrium constant in terms of concentration =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side= $n_{g,p}-n_{g.r}=1-2=-1$

$98.1=K_c(RT)^{-1}$

$98.1 =\frac{K_c}{RT}$

$K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}$

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

5 0

3 years ago

Read 2 more answers