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2 years ago

Can y’all please help me this is due tomorrow

Chemistry

2 answers:

Verizon [17]2 years ago

8 0

Answer:

1,crust

2, mesosphere

3,Inner core

4,Outer core

5, Mantle

Solnce55 [7]2 years ago

6 0

1. Crust
2. Mesosphere
5. Mantle
4. Outer core
3. Inner core

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Which medium slows the speed of light waves? Select all that apply.

77julia77 [94]

Answer : pool water

Explanation:

3 0

3 years ago

A 2.5 mol sample of phosphorus pentachloride, PCl5 dissociates at 160C and 1.00atm to give 0.338 mol of phosphorus trichloride a

Readme [11.4K]

Explanation:

Moles of phosphorus pentachloride present initially = 2.5 mol

Moles of phosphorus trichloride at equilibrium = 0.338 mol

$PCl_5\rightleftharpoons PCl_3+Cl_2$

Initially

2.5 mol 0 0

At equilibrium:

(2.5 - x) mol x x

So, from above, the moles of phosphorus trichloride at equilibrium , x= 0.338 mol

Mass of 0.338 moles of phosphorus trichloride at equilibrium:

= 0.338 mol × 137.5 g/mol = 46.475 g

Moles of phosphorus pentachloride present at equilibrium :

= (2.5 - 0.338) mol = 2.162 mol

Mass of 2.162 moles of phosphorus pentachloride at equilibrium:

= 2.162 mol × 208.5 g/mol = 450.777 g

Moles of chloride gas present at equilibrium : 0.338 mol

Mass of 0.338 moles of chloride gas at equilibrium:

= 0.338 mol × 71 g/mol = 23.998 g

3 0

3 years ago

A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t

jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for weak acids for the equilibria:

protonated lidocaine + H₂O ⇆ lidocaine + H₃O⁺

protonated lidocaine lidocaine H₃O⁺

Initial(M) 0.22 0 0

Change -x +x +x

Equilibrium 0.22 - x x x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] = x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸ = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22 - x ≈ 0.22

and solve for x. The pH will then be the negative log of this value.

8.71 x 10⁻⁷ = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0

3 years ago

15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 6.25 L. What is th

Y_Kistochka [10]

Answer:

The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

Explanation:

For N2,

Pressure(P₁)=125 kPa

Volume(V₁)=15·1 L

Temperature (T₁)=25°C=25+273 K=298 K

Similarly, for Oxygen,

Pressure(P₂)= 125 kPa

Volume(V₂)= 44.3 L

Temperature(T₂)=25°C= 298 K

Then, for the mixture,

Volumeof the mixture( V)= 6.25 L

Pressure(P)=?

Temperature (T)= 51°C = 51+273 K=324 K

Then, By Combined gas laws,

$\frac{P_{1} V_{1} }{T_{1} } +\frac{P_{2} V_{2} }{T_{2} } =\frac{PV}{T}$

or, $\frac{15.1*125}{298} +\frac{44.3*125}{298} =\frac{P*6.25}{324}$

or, $6.34+18.58=\frac{P*6.25}{324}$

or, $P=\frac{24.92*324}{6.25}$

∴P=1291.85 kPa

So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

3 0

2 years ago

What does an algae cell, tree,mushroom, and animal have in common

wel

Each one is a living organism.

5 0

3 years ago