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RoseWind [281]
3 years ago
5

Ethylene glycol (antifreeze) has a density of 1.11 g/cm3. What is the mass in grams of 387 mL of ethylene glycol?

Chemistry
1 answer:
harkovskaia [24]3 years ago
5 0
To start, 1 cubic centimeter = 1 milliliter, so now you have 1.11g/mL.

Now multiply 1.11 by 387 to get the mass of antifreeze in grams, since the mL is canceled out.

387 mL x 1.11g/mL = 429.57 g
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Answer : Option C) The Octet Rule

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In short, the tendency of an atom to fill its valence shell and attain a stable state it acquires or donates the electron is called as octet rule.

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2 years ago
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A solution contains 0.115 mol h2o and an unknown number of moles of nacl. the vapor pressure of the solution at 30°c is 25.7 tor
Valentin [98]
According to Raoult's low:
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∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr 
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∴ mole fraction of solvent = 25.7 / 31.8 =0.808 
when we assume the moles of solute NaCl = X 
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
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At the start of a reaction, there are 0.0249 mol N2,
gladu [14]

Answer:

Explanation:

The reaction is given as:

N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}

The reaction quotient is:

Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}

From the given information:

TO find each entity in the reaction quotient, we have:

[NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}

[N_2] = \dfrac{0.024 }{3.5}

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[H_2] = 9.17 \times 10^{-3}

∴

Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135

However; given that:

K_c = 1.2

By relating Q_c \ \ and  \ \ K_c, we will realize that Q_c \ \ <  \ \ K_c

The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.

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