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3 years ago

Which of the following is NOT an effect of human development on the Everglades?

Chemistry

2 answers:

Fofino [41]3 years ago

4 0

Answer: I think its A

Explanation:

Molodets [167]3 years ago

3 0

I’m pretty sure the answer is A

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I need help please!Perform the following conversion. Make sure your answer has the correct number of significant digits.

Nuetrik [128]

Answer:

3220000

Explanation:

1kg=10^6 Gg

3.22×10^6

5 0

3 years ago

Molarity formula what is the molarity of a solution in which 3 moles of nacl is dissolved in 1.5 kg of water

Igoryamba

Molarity is..
1.5 kg H₂O = 1.5L H₂O
So...
$\frac{moles}{volume} = molarity$
$\frac{3.0 mol}{1.5L} = molarity$
$2M = molarity$

7 0

3 years ago

Given the following information:Br2 bond energy = 193 kJ/mol F2 bond energy = 154 kJ/mol 1/2Br2(g) + 3/2F2(g) -> BrF3(g) = –3

arlik [135]

Answer:

C) 712 KJ/mol

Explanation:

ΔH°r = Σ Eb broken - Σ Eb formed

1/2Br2(g) + 3/2F2(g) → BrF3(g)

∴ ΔH°r = - 384 KJ/mol

∴ Br2 Eb = 193 KJ/mol

∴ F2 Eb = 154 KJ/mol

⇒ Σ Eb broken = (1/2)(Br-Br) + (3/2)(F-F)

⇒ Σ Eb broken = (1/2)(193 KJ/mol) + (3/2)(154 KJ/mol) = 327.5 KJ/mol

∴ Eb formed: Br-F

⇒ Σ Eb formed (Br-F) = Σ Eb broken - ΔH°r

⇒ Eb (Br-F) = 327.5 KJ/mol - ( - 384 KJ/mol )

⇒ Eb Br-F = 327.5 KJ/mol + 384 KJ/mol = 711.5 KJ/mol ≅ 712 KJ/mol

8 0

3 years ago

A pressure cooker contains 5.68 liters of air at a temperature of 394 K. If the absolute pressure of the air in the pressure coo

Arturiano [62]

Answer:

Explanation:

We shall find out volume of air at NTP or at 273 K and 10⁵ Pa ( 1 atm )

Let it be V₂

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

$\frac{2\times 10^5\times 5.68}{394} =\frac{10^5\times V_2}{273}$

V₂ = 7.87 litres

22.4 litres of any gas is equivalent to 1 mole

7.87 litres of air will be equivalent to

7.87 / 22.4 moles

= .35 moles .

7 0

3 years ago

Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

gtnhenbr [62]

Answer:

Approximately $1.10\; {\rm V}$ under standard conditions.

Explanation:

Equation for the overall reaction:

${\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s)$ .

Write down the ionic equation for this reaction:

$\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}$ .

The net ionic equation for this reaction would be:

${\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)$ .

In this reaction:

Zinc loses electrons and was oxidized (at the anode): ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ .
Copper gains electrons and was reduced (at the cathode): ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , is reduction and is likely on the table.

$E(\text{anode}) = -0.7618\; {\rm V}$ for ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , and
$E(\text{cathode}) = 0.3419\; {\rm V}$ for ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

The reduction potential of ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ would be $-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}$ , the opposite of the reverse reaction ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ .

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

$\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}$ .

7 0

2 years ago