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3 years ago

How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.88 ∘c?

1 answer:

6 0

The freezing point depression is calculated through the equation,
ΔT = (kf) x m
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
5.88 = (1.86)(m)
m is equal to 3.16m

Recall that molality is calculated through the equation,
molality = number of mols / kg of solvent
number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.

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N204(0) + 2NO2(g)

user100 [1]

setup 1 : to the right

setup 2 : equilibrium

setup 3 : to the left

<h3>Further explanation</h3>

The reaction quotient (Q) : determine a reaction has reached equilibrium

For reaction :

aA+bB⇔cC+dD

$\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}$

Comparing Q with K( the equilibrium constant) :

K is the product of ions in an equilibrium saturated state

Q is the product of the ion ions from the reacting substance

Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)

Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium

Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)

Keq = 6.16 x 10⁻³

Q for reaction N₂O₄(0) ⇒ 2NO₂(g)

$\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}$

Setup 1 :

$\tt Q=\dfrac{0.0064^2}{0.098}=0.000418=4.18\times 10^{-4}$

Q<K⇒The reaction moved to the right (products)

Setup 2 :

$\tt Q=\dfrac{0.0304^2}{0.15}=0.00616=6.16\times 10^{-3}$

Q=K⇒the system at equilibrium

Setup 3 :

$\tt Q=\dfrac{0.230^2}{0.420}=0.126$

Q>K⇒The reaction moved to the left (reactants)

8 0

3 years ago

Read 2 more answers

Refer to pic plssss

ira [324]

Answer:

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4 0

3 years ago

When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the ph

Scrat [10]

Answer:

E = 12.92 × 10^(-16) J

Explanation:

Formula for energy is;

E = hc/λ

Where;

h is Planck's constant = 6.63 x 10^(-34) J.s

c is speed of light = 3 × 10^(8) m/s

λ is wavelength = 0.154 nm = 0.154 × 10^(-9) m

E = (6.63 x 10^(-34) × 3 × 10^(8))/(0.154 × 10^(-9))

E = 12.92 × 10^(-16) J

6 0

3 years ago

Too much ___________ can slow and eventually stop the heart, and supplements should be taken only under the care of a physician.

Marina86 [1]

<span>Too much B. potassium can slow and eventually stop the heart, and supplements should be taken only under the care of a physician. </span>

6 0

3 years ago

The balanced equation below shows the products that are formed when pentane (C₅H₁₂) is combusted. <img src="https://tex.z-dn.net

masha68 [24]

Answer:

Option c → 8:1

Explanation:

This is the reaction:

C₅H₁₂ + 8O₂ → 10CO₂ + 6H₂O

1 mol of pentane needs 8 moles of oxygen to be combusted and this combustion produces 10 mol of carbon dioxide and 6 moles of water.

To determine the ratio, look the stoichiometry.

For every 8 moles of oxygen, I need 1 mole of pentane gas.

7 0

3 years ago