A drought is a long period of dryness. If animals fight during a drought, which basic need is most likely limited?

Explain interaction between sun system and hydrosphere

qwelly [4]

Both the Sun and the Earth are sources of heat that power an interconnected set of dynamic systems (lithosphere, hydrosphere & cryosphere, atmosphere, biosphere).

Within the Sun, heat is transferred by radiation and convection, which involves circulation of hydrogen ions. Within the Earth heat is transferred by conduction and convection, which involves circulation of silicates in the mantle and the crust, and by the circulation of iron in the liquid outer core. On the surface of the Earth and the atmosphere, heat emanating largely from the Sun is transferred by convection, which involving the circulation of water and carbon. Both the Sun and the Earth and their atmospheres are layered. Both systems evolve and change.

6 0

3 years ago

Osmosis occurs when two solutions of different concentrations are separated by a semipermeable membrane. The membrane allows sol

Finger [1]

Answer:

Part A → 7.82 atm

Part B → The unknown solution had the higher concentration

Part C → 0.83 mol/L

Explanation:

Part A

Osmotic pressure (π) = M . R. T . i

NaCl → Na⁺ + Cl⁻ (i =2)

0.923 g of NaCl must be dissolved in 100 mL of solution.

0.923 g / 58.45 g/m = 0.016 moles

Molarity is mol/L → 0.016 m / 0.1L = 0.16M

π = 0.16M . 0.08206 L.atm/molK . 298K . 2 ⇒ 7.82atm

Part. B

The solvent moves toward the solution of higher concentration (to dilute it) until the two solutions have the same concentration, or until gravity overtakes the osmotic pressure, Π. If the level of the unknown solution drops when it was connected to solution in part A, we can be sure that had a higher concentration.

Part. C

π = M . R . T

20.1 atm = M . 0.08206 L.atm/mol.K . 294K

20.1 atm / (0.08206 L.atm/mol.K . 294K) = 0.83 mol/L

8 0

3 years ago

A 0.366 mol sample of pcl5(g) is injected into an empty 4.45 l reaction vessel held at 250 °c. calculate the concentrations of p

Maru [420]

Chemical reaction: PCl₅ → PCl₃ + Cl₂.
n(PCl₅) = 0,366 mol.
V(PCl₅) = 4,45 L.
c(PCl₅) = n(PCl₅) ÷ V(PCl₅).
c(PCl₅) = 0,366 mol ÷ 4,45 L.
c(PCl₅) = 0,082 mol/L.
Kc = 1,80.
[PCl₃] = [Cl₂] = x.
Kc = [PCl₃] · [Cl₂] ÷ [PCl₅].
1,80 = x² ÷ (0,082 mol/L - x).
Solve quadratic eqaution: x = [PCl₃] = 0,078 mol/L.
[PCl₅] = 0,082 mol/L - 0,078 mol/L.
[PCl₅] = 0,004 mol/L.

6 0

3 years ago

Examine the equation.

Alisiya [41]

Explanation:

The reaction expression is given as;

2H₂ $_{g}$ + O₂ $_{g}$ → 2H₂O $_{l}$

From the balance reaction expression:

2 mole of hydrogen gas combines with 1 mole of oxygen gas on the reactant side;

This produces 2 mole of water on the product side of the expression.

The product is in liquid form.

This reaction is a synthesis reaction because a single product is formed from two reactants.

5 0

3 years ago

Consider the titration of a 20.0 mL sample of 0.500 M HCN (Ka =6.17x10-10) with 0.250 M KOH. a. (6pt) What is the initial pH? b.

Salsk061 [2.6K]

Answer:

a. pH = 4.75

b. pH = 9.20

c. pH = 8.42

d. pH = 13.53

Explanation:

This is a titration between a strong base, the KOH and a weak acid, HCN.

The initial pH is the pH, when you did not add the base yet, so it is the pH of the HCN

HCN + H2O ⇄ H₃O⁺ + CN⁻

Initial 0.5 - -

Eq. 0.5-x x x

Ka = x² / (0.5-x) = 6.17ₓ10⁻¹⁰

Ka is really small, so we can say that 0.5-x = 0.5. Then,

x² = 6.17ₓ10⁻¹⁰ . 0.5

x = √(6.17ₓ10⁻¹⁰ . 0.5) = 1.75×10⁻⁵ → [H₃O⁺]

pH = - log [H₃O⁺] → - log 1.75×10⁻⁵ = 4.75

b. First of all, we determine the moles of base, we are adding.

0.250 mol/L . 0.006 L = 0.0015 moles

In conclussion we have 0.0015 moles of OH⁻

Now, we determine the moles of our acid.

0.500 mol/L . 0.020L = 0.01 moles

The 0.0015 moles of OH⁻ will be neutralized with the acid, so:

HCN + OH⁻ → H₂O + CN⁻

0.01 0.0015 0.0085

The hydroxides are neutralized with the proton from the weak acid, so we have 0.0085 moles of cyanide and 0.0085 moles of HCN. (0.01-0.0015)

Our new volume is 20 mL and 6mL that we added, so, 26mL

This is a buffer with the weak acid, and its conjugate base.

Our concentrations are 0.0085 moles / 0.026 L = 0.327 M

We apply Henderson-Hasselbach

pH = pKa + log (base/acid) → pH = 9.20 + log (0.327/0.327)

pH = pKa

c. When we add 40 mL, our volume is 20mL +40mL = 60 mL

These are the moles, we add:

0.040 L . 0.250 mol/L = 0.01 moles of KOH (moles of OH⁻)

HCN + OH⁻ → H₂O + CN⁻

0.01 0.01 0.01

All the hydroxides have neutralized all the moles from the HCN, so we only have in solution, cyanhide. This is the equivalence point.

0.01 moles / 0.060 L = 0.16 M → [CN⁻]

pH at this point will be

CN⁻ + H₂O ⇄ HCN + OH⁻ Kb = 1.62ₓ10⁻⁵ (Kw/Ka)

In. 0.16 - -

Eq. 0.16-x x x

Kb = x² / (0.16-x)

We can also assume that 0.16-x = 0.16. Then:

[OH⁻] = √(Kb . 0.16) → √(1.62ₓ10⁻⁵ . 0.16) = 2.59×10⁻⁶

- log [OH⁻] = pOH → - log 2.59×10⁻⁶ = 5.58

pH = 14 - pOH → 14 - 5.58 = 8.42

This is a basic pH, because the titration is between a weak acid and a strong base.

d. When we add 42 mL of base, our volume is 20mL + 42 mL = 62 mL

We add 0.5 mol/L . 0.062L = 0.031 moles

These are the moles of OH⁻ , so as we have neutralized all the acid with 40 mL, with 42 mL of base, we only have base in solution.

0.031 moles - 0.01 moles = 0.021 moles of OH⁻

[OH⁻] = 0.021 moles / 0.062L = 0.34M

- log [OH⁻] = pOH → - log 0.34 = 0.47

pH = 14-pH → 14 - 0.47 = 13.53

8 0

3 years ago